4x^2+x=96

Solution for 4x^2+x=96 equation:

4x^2+x=96

We move all terms to the left:

4x^2+x-(96)=0

a = 4; b = 1; c = -96;
Δ = b2-4ac
Δ = 12-4·4·(-96)
Δ = 1537
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{-(1)-\sqrt{1537}}{2*4}=\frac{-1-\sqrt{1537}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1537}}{2*4}=\frac{-1+\sqrt{1537}}{8}
$