8x^2+40x+18=0

Solution for 8x^2+40x+18=0 equation:

8x^2+40x+18=0

a = 8; b = 40; c = +18;
Δ = b2-4ac
Δ = 402-4·8·18
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-32}{2*8}=\frac{-72}{16}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+32}{2*8}=\frac{-8}{16}
=-1/2
$