16x^2+12x-81=0

Solution for 16x^2+12x-81=0 equation:

16x^2+12x-81=0

a = 16; b = 12; c = -81;
Δ = b2-4ac
Δ = 122-4·16·(-81)
Δ = 5328
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5328}=\sqrt{144*37}=\sqrt{144}*\sqrt{37}=12\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{37}}{2*16}=\frac{-12-12\sqrt{37}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{37}}{2*16}=\frac{-12+12\sqrt{37}}{32}
$