5^2+25+3=7x^2+10

Solution for 5^2+25+3=7x^2+10 equation:

5^2+25+3=7x^2+10

We move all terms to the left:

5^2+25+3-(7x^2+10)=0

We add all the numbers together, and all the variables

-(7x^2+10)+53=0

We get rid of parentheses

-7x^2-10+53=0

We add all the numbers together, and all the variables

-7x^2+43=0

a = -7; b = 0; c = +43;
Δ = b2-4ac
Δ = 02-4·(-7)·43
Δ = 1204
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:

$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{301}}{2*-7}=\frac{0-2\sqrt{301}}{-14}
=-\frac{2\sqrt{301}}{-14}
=-\frac{\sqrt{301}}{-7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{301}}{2*-7}=\frac{0+2\sqrt{301}}{-14}
=\frac{2\sqrt{301}}{-14}
=\frac{\sqrt{301}}{-7}
$