(5r-2)=(r^2-38)

Solution for (5r-2)=(r^2-38) equation:

(5r-2)=(r^2-38)

We move all terms to the left:

(5r-2)-((r^2-38))=0

We get rid of parentheses

5r-((r^2-38))-2=0


We calculate terms in parentheses: -((r^2-38)), so:

(r^2-38)

We get rid of parentheses

r^2-38

Back to the equation:

-(r^2-38)


We get rid of parentheses

-r^2+5r+38-2=0

We add all the numbers together, and all the variables

-1r^2+5r+36=0

a = -1; b = 5; c = +36;
Δ = b2-4ac
Δ = 52-4·(-1)·36
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*-1}=\frac{-18}{-2}
=+9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-1}=\frac{8}{-2}
=-4
$