4x2-68=5x2+21x

Solution for 4x2-68=5x2+21x equation:

4x^2-68=5x^2+21x

We move all terms to the left:

4x^2-68-(5x^2+21x)=0

We get rid of parentheses

4x^2-5x^2-21x-68=0

We add all the numbers together, and all the variables

-1x^2-21x-68=0

a = -1; b = -21; c = -68;
Δ = b2-4ac
Δ = -212-4·(-1)·(-68)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-13}{2*-1}=\frac{8}{-2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+13}{2*-1}=\frac{34}{-2}
=-17
$