3x^2+64x-63=0

Solution for 3x^2+64x-63=0 equation:

3x^2+64x-63=0

a = 3; b = 64; c = -63;
Δ = b2-4ac
Δ = 642-4·3·(-63)
Δ = 4852
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:

$\sqrt{\Delta}=\sqrt{4852}=\sqrt{4*1213}=\sqrt{4}*\sqrt{1213}=2\sqrt{1213}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-2\sqrt{1213}}{2*3}=\frac{-64-2\sqrt{1213}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+2\sqrt{1213}}{2*3}=\frac{-64+2\sqrt{1213}}{6}
$