4a^2+a=68

Solution for 4a^2+a=68 equation:

4a^2+a=68

We move all terms to the left:

4a^2+a-(68)=0

a = 4; b = 1; c = -68;
Δ = b2-4ac
Δ = 12-4·4·(-68)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-33}{2*4}=\frac{-34}{8}
=-4+1/4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+33}{2*4}=\frac{32}{8}
=4
$