(10r2-79r+63)=(r-7)=

Solution for (10r2-79r+63)=(r-7)= equation:

(10r^2-79r+63)=(r-7)=

We move all terms to the left:

(10r^2-79r+63)-((r-7))=0

We get rid of parentheses

10r^2-79r-((r-7))+63=0


We calculate terms in parentheses: -((r-7)), so:

(r-7)

We get rid of parentheses

r-7

Back to the equation:

-(r-7)


We get rid of parentheses

10r^2-79r-r+7+63=0

We add all the numbers together, and all the variables

10r^2-80r+70=0

a = 10; b = -80; c = +70;
Δ = b2-4ac
Δ = -802-4·10·70
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-60}{2*10}=\frac{20}{20}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+60}{2*10}=\frac{140}{20}
=7
$