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 = 817
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{817}}{2*2}=\frac{21-\sqrt{817}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{817}}{2*2}=\frac{21+\sqrt{817}}{4}
$