-42r2+12r+86=0

Solution for -42r2+12r+86=0 equation:

-42r^2+12r+86=0

a = -42; b = 12; c = +86;
Δ = b2-4ac
Δ = 122-4·(-42)·86
Δ = 14592
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:

$\sqrt{\Delta}=\sqrt{14592}=\sqrt{256*57}=\sqrt{256}*\sqrt{57}=16\sqrt{57}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16\sqrt{57}}{2*-42}=\frac{-12-16\sqrt{57}}{-84}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16\sqrt{57}}{2*-42}=\frac{-12+16\sqrt{57}}{-84}
$