Solve the quadratic equation $25x^2+2x+5=0$

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Final answer to the problem

$x=\frac{-2+\sqrt{496}i}{50},\:x=\frac{-2-\sqrt{496}i}{50}$
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Step-by-step Solution

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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 $a=25$, $b=2$ and $c=5$. Then substitute the values of the coefficients of the equation in the quadratic formula: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-2\pm \sqrt{2^2-4\cdot 25\cdot 5}}{2\cdot 25}$

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$x=\frac{-2\pm \sqrt{2^2-4\cdot 25\cdot 5}}{2\cdot 25}$

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Learn how to solve problems step by step online. Solve the quadratic equation 25x^2+2x+5=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=25, b=2 and c=5. 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 (-). Combining all solutions, the 2 solutions of the equation are.

Final answer to the problem

$x=\frac{-2+\sqrt{496}i}{50},\:x=\frac{-2-\sqrt{496}i}{50}$

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Function Plot

Plotting: $25x^2+2x+5$

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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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

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