4r+3-1=2(2r+2)-r2

Solution for 4r+3-1=2(2r+2)-r2 equation:

4r+3-1=2(2r+2)-r2

We move all terms to the left:

4r+3-1-(2(2r+2)-r2)=0

We add all the numbers together, and all the variables

4r-(2(2r+2)-r2)+2=0


We calculate terms in parentheses: -(2(2r+2)-r2), so:

2(2r+2)-r2

We add all the numbers together, and all the variables

-1r^2+2(2r+2)

We multiply parentheses

-1r^2+4r+4

Back to the equation:

-(-1r^2+4r+4)


We get rid of parentheses

1r^2-4r+4r-4+2=0

We add all the numbers together, and all the variables

r^2-2=0

a = 1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·1·(-2)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*1}=\frac{0-2\sqrt{2}}{2}
=-\frac{2\sqrt{2}}{2}
=-\sqrt{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*1}=\frac{0+2\sqrt{2}}{2}
=\frac{2\sqrt{2}}{2}
=\sqrt{2}
$