-7r^2-12r+4=0

Solution for -7r^2-12r+4=0 equation:

-7r^2-12r+4=0

a = -7; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·(-7)·4
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*-7}=\frac{-4}{-14}
=2/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*-7}=\frac{28}{-14}
=-2
$