10x^2=-4x+23

Solution for 10x^2=-4x+23 equation:

10x^2=-4x+23

We move all terms to the left:

10x^2-(-4x+23)=0

We get rid of parentheses

10x^2+4x-23=0

a = 10; b = 4; c = -23;
Δ = b2-4ac
Δ = 42-4·10·(-23)
Δ = 936
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:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{26}}{2*10}=\frac{-4-6\sqrt{26}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{26}}{2*10}=\frac{-4+6\sqrt{26}}{20}
$