3x^2+10x=63+8x

Solution for 3x^2+10x=63+8x equation:

3x^2+10x=63+8x

We move all terms to the left:

3x^2+10x-(63+8x)=0

We add all the numbers together, and all the variables

3x^2+10x-(8x+63)=0

We get rid of parentheses

3x^2+10x-8x-63=0

We add all the numbers together, and all the variables

3x^2+2x-63=0

a = 3; b = 2; c = -63;
Δ = b2-4ac
Δ = 22-4·3·(-63)
Δ = 760
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{190}}{2*3}=\frac{-2-2\sqrt{190}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{190}}{2*3}=\frac{-2+2\sqrt{190}}{6}
$