3x^2+21x=294

Solution for 3x^2+21x=294 equation:

3x^2+21x=294

We move all terms to the left:

3x^2+21x-(294)=0

a = 3; b = 21; c = -294;
Δ = b2-4ac
Δ = 212-4·3·(-294)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-63}{2*3}=\frac{-84}{6}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+63}{2*3}=\frac{42}{6}
=7
$