3r^2-5(r+1)=7r+5

Solution for 3r^2-5(r+1)=7r+5 equation:

3r^2-5(r+1)=7r+5

We move all terms to the left:

3r^2-5(r+1)-(7r+5)=0

We multiply parentheses

3r^2-5r-(7r+5)-5=0

We get rid of parentheses

3r^2-5r-7r-5-5=0

We add all the numbers together, and all the variables

3r^2-12r-10=0

a = 3; b = -12; c = -10;
Δ = b2-4ac
Δ = -122-4·3·(-10)
Δ = 264
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{66}}{2*3}=\frac{12-2\sqrt{66}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{66}}{2*3}=\frac{12+2\sqrt{66}}{6}
$