2r^2+12r-32=0

Solution for 2r^2+12r-32=0 equation:

2r^2+12r-32=0

a = 2; b = 12; c = -32;
Δ = b2-4ac
Δ = 122-4·2·(-32)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*2}=\frac{-32}{4}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*2}=\frac{8}{4}
=2
$