3r-5=2(4r-9)r=

Solution for 3r-5=2(4r-9)r= equation:

3r-5=2(4r-9)r=

We move all terms to the left:

3r-5-(2(4r-9)r)=0


We calculate terms in parentheses: -(2(4r-9)r), so:

2(4r-9)r

We multiply parentheses

8r^2-18r

Back to the equation:

-(8r^2-18r)


We get rid of parentheses

-8r^2+3r+18r-5=0

We add all the numbers together, and all the variables

-8r^2+21r-5=0

a = -8; b = 21; c = -5;
Δ = b2-4ac
Δ = 212-4·(-8)·(-5)
Δ = 281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{281}}{2*-8}=\frac{-21-\sqrt{281}}{-16}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{281}}{2*-8}=\frac{-21+\sqrt{281}}{-16}
$