2x^2+40=21x

Solution for 2x^2+40=21x equation:

2x^2+40=21x

We move all terms to the left:

2x^2+40-(21x)=0

a = 2; b = -21; c = +40;
Δ = b2-4ac
Δ = -212-4·2·40
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-11}{2*2}=\frac{10}{4}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+11}{2*2}=\frac{32}{4}
=8
$