3x^2+10x=18

Solution for 3x^2+10x=18 equation:

3x^2+10x=18

We move all terms to the left:

3x^2+10x-(18)=0

a = 3; b = 10; c = -18;
Δ = b2-4ac
Δ = 102-4·3·(-18)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{79}}{2*3}=\frac{-10-2\sqrt{79}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{79}}{2*3}=\frac{-10+2\sqrt{79}}{6}
$