10x+40=x^2+7

Solution for 10x+40=x^2+7 equation:

10x+40=x^2+7

We move all terms to the left:

10x+40-(x^2+7)=0

We get rid of parentheses

-x^2+10x-7+40=0

We add all the numbers together, and all the variables

-1x^2+10x+33=0

a = -1; b = 10; c = +33;
Δ = b2-4ac
Δ = 102-4·(-1)·33
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{58}}{2*-1}=\frac{-10-2\sqrt{58}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{58}}{2*-1}=\frac{-10+2\sqrt{58}}{-2}
$