8x^2+4x=21

Solution for 8x^2+4x=21 equation:

8x^2+4x=21

We move all terms to the left:

8x^2+4x-(21)=0

a = 8; b = 4; c = -21;
Δ = b2-4ac
Δ = 42-4·8·(-21)
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{43}}{2*8}=\frac{-4-4\sqrt{43}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{43}}{2*8}=\frac{-4+4\sqrt{43}}{16}
$