4x^2+7x=320

Solution for 4x^2+7x=320 equation:

4x^2+7x=320

We move all terms to the left:

4x^2+7x-(320)=0

a = 4; b = 7; c = -320;
Δ = b2-4ac
Δ = 72-4·4·(-320)
Δ = 5169
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{5169}}{2*4}=\frac{-7-\sqrt{5169}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{5169}}{2*4}=\frac{-7+\sqrt{5169}}{8}
$