4a^2+8a=140

Solution for 4a^2+8a=140 equation:

4a^2+8a=140

We move all terms to the left:

4a^2+8a-(140)=0

a = 4; b = 8; c = -140;
Δ = b2-4ac
Δ = 82-4·4·(-140)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-48}{2*4}=\frac{-56}{8}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+48}{2*4}=\frac{40}{8}
=5
$