16x^2-67x=20

Solution for 16x^2-67x=20 equation:

16x^2-67x=20

We move all terms to the left:

16x^2-67x-(20)=0

a = 16; b = -67; c = -20;
Δ = b2-4ac
Δ = -672-4·16·(-20)
Δ = 5769
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:

$\sqrt{\Delta}=\sqrt{5769}=\sqrt{9*641}=\sqrt{9}*\sqrt{641}=3\sqrt{641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-3\sqrt{641}}{2*16}=\frac{67-3\sqrt{641}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+3\sqrt{641}}{2*16}=\frac{67+3\sqrt{641}}{32}
$