8r^2+16r-12=0

Solution for 8r^2+16r-12=0 equation:

8r^2+16r-12=0

a = 8; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·8·(-12)
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{10}}{2*8}=\frac{-16-8\sqrt{10}}{16}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{10}}{2*8}=\frac{-16+8\sqrt{10}}{16}
$