25x^2-62+36=0

Solution for 25x^2-62+36=0 equation:

25x^2-62+36=0

We add all the numbers together, and all the variables

25x^2-26=0

a = 25; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·25·(-26)
Δ = 2600
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:

$\sqrt{\Delta}=\sqrt{2600}=\sqrt{100*26}=\sqrt{100}*\sqrt{26}=10\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{26}}{2*25}=\frac{0-10\sqrt{26}}{50}
=-\frac{10\sqrt{26}}{50}
=-\frac{\sqrt{26}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{26}}{2*25}=\frac{0+10\sqrt{26}}{50}
=\frac{10\sqrt{26}}{50}
=\frac{\sqrt{26}}{5}
$