6x^2+10x=3x+20

Solution for 6x^2+10x=3x+20 equation:

6x^2+10x=3x+20

We move all terms to the left:

6x^2+10x-(3x+20)=0

We get rid of parentheses

6x^2+10x-3x-20=0

We add all the numbers together, and all the variables

6x^2+7x-20=0

a = 6; b = 7; c = -20;
Δ = b2-4ac
Δ = 72-4·6·(-20)
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{-(7)-23}{2*6}=\frac{-30}{12}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*6}=\frac{16}{12}
=1+1/3
$