3r-7=2(4-3r)r

Solution for 3r-7=2(4-3r)r equation:

3r-7=2(4-3r)r

We move all terms to the left:

3r-7-(2(4-3r)r)=0

We add all the numbers together, and all the variables

3r-(2(-3r+4)r)-7=0


We calculate terms in parentheses: -(2(-3r+4)r), so:

2(-3r+4)r

We multiply parentheses

-6r^2+8r

Back to the equation:

-(-6r^2+8r)


We get rid of parentheses

6r^2-8r+3r-7=0

We add all the numbers together, and all the variables

6r^2-5r-7=0

a = 6; b = -5; c = -7;
Δ = b2-4ac
Δ = -52-4·6·(-7)
Δ = 193
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{-(-5)-\sqrt{193}}{2*6}=\frac{5-\sqrt{193}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{193}}{2*6}=\frac{5+\sqrt{193}}{12}
$