7x^2+2x-8=0

Solution for 7x^2+2x-8=0 equation:

7x^2+2x-8=0

a = 7; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·7·(-8)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{57}}{2*7}=\frac{-2-2\sqrt{57}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{57}}{2*7}=\frac{-2+2\sqrt{57}}{14}
$