3x^2+19x=40

Solution for 3x^2+19x=40 equation:

3x^2+19x=40

We move all terms to the left:

3x^2+19x-(40)=0

a = 3; b = 19; c = -40;
Δ = b2-4ac
Δ = 192-4·3·(-40)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-29}{2*3}=\frac{-48}{6}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+29}{2*3}=\frac{10}{6}
=1+2/3
$