4=r^2+4r+7

Solution for 4=r^2+4r+7 equation:

4=r^2+4r+7

We move all terms to the left:

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

We get rid of parentheses

-r^2-4r-7+4=0

We add all the numbers together, and all the variables

-1r^2-4r-3=0

a = -1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·(-1)·(-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{-(-4)-2}{2*-1}=\frac{2}{-2}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*-1}=\frac{6}{-2}
=-3
$