21x^2+10x=5x+6

Solution for 21x^2+10x=5x+6 equation:

21x^2+10x=5x+6

We move all terms to the left:

21x^2+10x-(5x+6)=0

We get rid of parentheses

21x^2+10x-5x-6=0

We add all the numbers together, and all the variables

21x^2+5x-6=0

a = 21; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·21·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-23}{2*21}=\frac{-28}{42}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+23}{2*21}=\frac{18}{42}
=3/7
$