(r-8)(2r=5)

Solution for (r-8)(2r=5) equation:

(r-8)(2r=5)

We move all terms to the left:

(r-8)(2r-(5))=0

We multiply parentheses ..

(+2r^2-5r-16r+40)=0

We get rid of parentheses

2r^2-5r-16r+40=0

We add all the numbers together, and all the variables

2r^2-21r+40=0

a = 2; b = -21; c = +40;
Δ = b2-4ac
Δ = -212-4·2·40
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-11}{2*2}=\frac{10}{4}
=2+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+11}{2*2}=\frac{32}{4}
=8
$