8x^2+40x+35=3

Solution for 8x^2+40x+35=3 equation:

8x^2+40x+35=3

We move all terms to the left:

8x^2+40x+35-(3)=0

We add all the numbers together, and all the variables

8x^2+40x+32=0

a = 8; b = 40; c = +32;
Δ = b2-4ac
Δ = 402-4·8·32
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24}{2*8}=\frac{-64}{16}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24}{2*8}=\frac{-16}{16}
=-1
$