(7^3-5+2)-(5x^2+5)=0

Solution for (7^3-5+2)-(5x^2+5)=0 equation:

(7^3-5+2)-(5x^2+5)=0

We add all the numbers together, and all the variables

-(5x^2+5)+340=0

We get rid of parentheses

-5x^2-5+340=0

We add all the numbers together, and all the variables

-5x^2+335=0

a = -5; b = 0; c = +335;
Δ = b2-4ac
Δ = 02-4·(-5)·335
Δ = 6700
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:

$\sqrt{\Delta}=\sqrt{6700}=\sqrt{100*67}=\sqrt{100}*\sqrt{67}=10\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{67}}{2*-5}=\frac{0-10\sqrt{67}}{-10}
=-\frac{10\sqrt{67}}{-10}
=-\frac{\sqrt{67}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{67}}{2*-5}=\frac{0+10\sqrt{67}}{-10}
=\frac{10\sqrt{67}}{-10}
=\frac{\sqrt{67}}{-1}
$