(5r+3)(r+1)=)

Solution for (5r+3)(r+1)=) equation:

(5r+3)(r+1)=)

We move all terms to the left:

(5r+3)(r+1)-())=0

We add all the numbers together, and all the variables

(5r+3)(r+1)=0

We multiply parentheses ..

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

We get rid of parentheses

5r^2+5r+3r+3=0

We add all the numbers together, and all the variables

5r^2+8r+3=0

a = 5; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·5·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*5}=\frac{-10}{10}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*5}=\frac{-6}{10}
=-3/5
$