23x^2-98x+89=0

Solution for 23x^2-98x+89=0 equation:

23x^2-98x+89=0

a = 23; b = -98; c = +89;
Δ = b2-4ac
Δ = -982-4·23·89
Δ = 1416
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:

$\sqrt{\Delta}=\sqrt{1416}=\sqrt{4*354}=\sqrt{4}*\sqrt{354}=2\sqrt{354}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-2\sqrt{354}}{2*23}=\frac{98-2\sqrt{354}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+2\sqrt{354}}{2*23}=\frac{98+2\sqrt{354}}{46}
$