10x+48=x2

Solution for 10x+48=x2 equation:

10x+48=x2

We move all terms to the left:

10x+48-(x2)=0

We add all the numbers together, and all the variables

-1x^2+10x+48=0

a = -1; b = 10; c = +48;
Δ = b2-4ac
Δ = 102-4·(-1)·48
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*-1}=\frac{-10-2\sqrt{73}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*-1}=\frac{-10+2\sqrt{73}}{-2}
$