26x=8x^2+21

Solution for 26x=8x^2+21 equation:

26x=8x^2+21

We move all terms to the left:

26x-(8x^2+21)=0

We get rid of parentheses

-8x^2+26x-21=0

a = -8; b = 26; c = -21;
Δ = b2-4ac
Δ = 262-4·(-8)·(-21)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2}{2*-8}=\frac{-28}{-16}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2}{2*-8}=\frac{-24}{-16}
=1+1/2
$