7x^2+40x+12=0

Solution for 7x^2+40x+12=0 equation:

7x^2+40x+12=0

a = 7; b = 40; c = +12;
Δ = b2-4ac
Δ = 402-4·7·12
Δ = 1264
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:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{79}}{2*7}=\frac{-40-4\sqrt{79}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{79}}{2*7}=\frac{-40+4\sqrt{79}}{14}
$