4r^2+8r-3=0

Solution for 4r^2+8r-3=0 equation:

4r^2+8r-3=0

a = 4; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·4·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{7}}{2*4}=\frac{-8-4\sqrt{7}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{7}}{2*4}=\frac{-8+4\sqrt{7}}{8}
$