68x+4=21x^2

Solution for 68x+4=21x^2 equation:

68x+4=21x^2

We move all terms to the left:

68x+4-(21x^2)=0

determiningTheFunctionDomain
-21x^2+68x+4=0

a = -21; b = 68; c = +4;
Δ = b2-4ac
Δ = 682-4·(-21)·4
Δ = 4960
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4960}=\sqrt{16*310}=\sqrt{16}*\sqrt{310}=4\sqrt{310}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4\sqrt{310}}{2*-21}=\frac{-68-4\sqrt{310}}{-42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4\sqrt{310}}{2*-21}=\frac{-68+4\sqrt{310}}{-42}
$