40=10x+3x^2

Solution for 40=10x+3x^2 equation:

40=10x+3x^2

We move all terms to the left:

40-(10x+3x^2)=0

We get rid of parentheses

-3x^2-10x+40=0

a = -3; b = -10; c = +40;
Δ = b2-4ac
Δ = -102-4·(-3)·40
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{145}}{2*-3}=\frac{10-2\sqrt{145}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{145}}{2*-3}=\frac{10+2\sqrt{145}}{-6}
$