7x2+20x+8=0

Solution for 7x2+20x+8=0 equation:

7x^2+20x+8=0

a = 7; b = 20; c = +8;
Δ = b2-4ac
Δ = 202-4·7·8
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{11}}{2*7}=\frac{-20-4\sqrt{11}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{11}}{2*7}=\frac{-20+4\sqrt{11}}{14}
$