10r2-21r=-4r+6

Solution for 10r2-21r=-4r+6 equation:

10r^2-21r=-4r+6

We move all terms to the left:

10r^2-21r-(-4r+6)=0

We get rid of parentheses

10r^2-21r+4r-6=0

We add all the numbers together, and all the variables

10r^2-17r-6=0

a = 10; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·10·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-23}{2*10}=\frac{-6}{20}
=-3/10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+23}{2*10}=\frac{40}{20}
=2
$