(2x)^2+7x=79

Solution for (2x)^2+7x=79 equation:

(2x)^2+7x=79

We move all terms to the left:

(2x)^2+7x-(79)=0

a = 2; b = 7; c = -79;
Δ = b2-4ac
Δ = 72-4·2·(-79)
Δ = 681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{681}}{2*2}=\frac{-7-\sqrt{681}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{681}}{2*2}=\frac{-7+\sqrt{681}}{4}
$