2x^2+40x+78=0

Solution for 2x^2+40x+78=0 equation:

2x^2+40x+78=0

a = 2; b = 40; c = +78;
Δ = b2-4ac
Δ = 402-4·2·78
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{61}}{2*2}=\frac{-40-4\sqrt{61}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{61}}{2*2}=\frac{-40+4\sqrt{61}}{4}
$