16x^2+72x=65

Solution for 16x^2+72x=65 equation:

16x^2+72x=65

We move all terms to the left:

16x^2+72x-(65)=0

a = 16; b = 72; c = -65;
Δ = b2-4ac
Δ = 722-4·16·(-65)
Δ = 9344
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:

$\sqrt{\Delta}=\sqrt{9344}=\sqrt{64*146}=\sqrt{64}*\sqrt{146}=8\sqrt{146}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{146}}{2*16}=\frac{-72-8\sqrt{146}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{146}}{2*16}=\frac{-72+8\sqrt{146}}{32}
$