8r^2+16r-15=0

Solution for 8r^2+16r-15=0 equation:

8r^2+16r-15=0

a = 8; b = 16; c = -15;
Δ = b2-4ac
Δ = 162-4·8·(-15)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{46}}{2*8}=\frac{-16-4\sqrt{46}}{16}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{46}}{2*8}=\frac{-16+4\sqrt{46}}{16}
$