2x^2=21x-47

Solution for 2x^2=21x-47 equation:

2x^2=21x-47

We move all terms to the left:

2x^2-(21x-47)=0

We get rid of parentheses

2x^2-21x+47=0

a = 2; b = -21; c = +47;
Δ = b2-4ac
Δ = -212-4·2·47
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{65}}{2*2}=\frac{21-\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{65}}{2*2}=\frac{21+\sqrt{65}}{4}
$