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# Solve the quadratic equation $15x^2-26x-57=0$

## Step-by-step Solution

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###  Videos

$x=3,\:x=-1.266667$
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##  Step-by-step Solution 

Problem to solve:

$15x^2-26x-57=0$

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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 $a=15$, $b=-26$ and $c=-57$. Then substitute the values of the coefficients of the equation in the quadratic formula: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{26\pm \sqrt{{\left(-26\right)}^2-4\cdot 15\cdot -57}}{2\cdot 15}$

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

$x=\frac{26\pm \sqrt{{\left(-26\right)}^2-4\cdot 15\cdot -57}}{2\cdot 15}$

Learn how to solve quadratic equations problems step by step online. Solve the quadratic equation 15x^2-26x-57=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=15, b=-26 and c=-57. Then substitute the values of the coefficients of the equation in the quadratic formula: \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Simplify \frac{26\pm \sqrt{{\left(-26\right)}^2-4\cdot 15\cdot -57}}{2\cdot 15}. 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 26 and -64.

$x=3,\:x=-1.266667$

##  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

Solve for xFind the rootsSolve by factoringSolve by completing the squareSolve by quadratic formulaFind break even points

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d/dx
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sin
cos
tan
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sec
csc

asin
acos
atan
acot
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acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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