-4r^2+2r+12=0

Solution for -4r^2+2r+12=0 equation:

-4r^2+2r+12=0

a = -4; b = 2; c = +12;
Δ = b2-4ac
Δ = 22-4·(-4)·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*-4}=\frac{-16}{-8}
=+2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*-4}=\frac{12}{-8}
=-1+1/2
$