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Solve the quadratic equation $5x^2-160x+1200=0$

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 Final answer to the problem

$x=20,\:x=12$
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 Step-by-step Solution 

How should I solve this problem?

• Choose an option
• Solve for x
• Find the derivative using the definition
• Solve by quadratic formula (general formula)
• Simplify
• Find the integral
• Find the derivative
• Factor
• Factor by completing the square
• Find the roots
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To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where in this case $a=5$, $b=-160$ and $c=1200$. Then substitute the values of the coefficients of the equation in the quadratic formula: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{160\pm \sqrt{{\left(-160\right)}^2-4\cdot 5\cdot 1200}}{2\cdot 5}$

Learn how to solve quadratic equations problems step by step online.

$x=\frac{160\pm \sqrt{{\left(-160\right)}^2-4\cdot 5\cdot 1200}}{2\cdot 5}$

Learn how to solve quadratic equations problems step by step online. Solve the quadratic equation 5x^2-160x+1200=0. To find the roots of a polynomial of the form ax^2+bx+c we use the quadratic formula, where in this case a=5, b=-160 and c=1200. Then substitute the values of the coefficients of the equation in the quadratic formula: \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Simplifying. To obtain the two solutions, divide the equation in two equations, one when \pm is positive (+), and another when \pm is negative (-). Subtract the values 160 and -40.

 Final answer to the problem

$x=20,\:x=12$

 Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

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9
0
a
b
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g
m
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u
v
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x
y
z
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+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

The quadratic equations (or second degree equations) are those equations where the greatest exponent to which the unknown is raised is the exponent 2.