12r2+28r-80=0

Solution for 12r2+28r-80=0 equation:

12r^2+28r-80=0

a = 12; b = 28; c = -80;
Δ = b2-4ac
Δ = 282-4·12·(-80)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-68}{2*12}=\frac{-96}{24}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+68}{2*12}=\frac{40}{24}
=1+2/3
$