2r^2+23=14r

Solution for 2r^2+23=14r equation:

2r^2+23=14r

We move all terms to the left:

2r^2+23-(14r)=0

a = 2; b = -14; c = +23;
Δ = b2-4ac
Δ = -142-4·2·23
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{3}}{2*2}=\frac{14-2\sqrt{3}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{3}}{2*2}=\frac{14+2\sqrt{3}}{4}
$