3x^2+6x+8=-x^2+21x-6

Solution for 3x^2+6x+8=-x^2+21x-6 equation:

3x^2+6x+8=-x^2+21x-6

We move all terms to the left:

3x^2+6x+8-(-x^2+21x-6)=0

We get rid of parentheses

3x^2+x^2-21x+6x+6+8=0

We add all the numbers together, and all the variables

4x^2-15x+14=0

a = 4; b = -15; c = +14;
Δ = b2-4ac
Δ = -152-4·4·14
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-1}{2*4}=\frac{14}{8}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+1}{2*4}=\frac{16}{8}
=2
$