3x2+7x=40

Solution for 3x2+7x=40 equation:

3x^2+7x=40

We move all terms to the left:

3x^2+7x-(40)=0

a = 3; b = 7; c = -40;
Δ = b2-4ac
Δ = 72-4·3·(-40)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*3}=\frac{16}{6}
=2+2/3
$