3x^2+21x=18

Solution for 3x^2+21x=18 equation:

3x^2+21x=18

We move all terms to the left:

3x^2+21x-(18)=0

a = 3; b = 21; c = -18;
Δ = b2-4ac
Δ = 212-4·3·(-18)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{73}}{2*3}=\frac{-21-3\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{73}}{2*3}=\frac{-21+3\sqrt{73}}{6}
$