320=(x)(4x+7)

Solution for 320=(x)(4x+7) equation:

320=(x)(4x+7)

We move all terms to the left:

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


We calculate terms in parentheses: -(x(4x+7)), so:

x(4x+7)

We multiply parentheses

4x^2+7x

Back to the equation:

-(4x^2+7x)


We get rid of parentheses

-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}
$