3x^2+10x=44

Solution for 3x^2+10x=44 equation:

3x^2+10x=44

We move all terms to the left:

3x^2+10x-(44)=0

a = 3; b = 10; c = -44;
Δ = b2-4ac
Δ = 102-4·3·(-44)
Δ = 628
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:

$\sqrt{\Delta}=\sqrt{628}=\sqrt{4*157}=\sqrt{4}*\sqrt{157}=2\sqrt{157}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{157}}{2*3}=\frac{-10-2\sqrt{157}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{157}}{2*3}=\frac{-10+2\sqrt{157}}{6}
$