4r+r^2=33/4

Solution for 4r+r^2=33/4 equation:

4r+r^2=33/4

We move all terms to the left:

4r+r^2-(33/4)=0

We add all the numbers together, and all the variables

r^2+4r-(+33/4)=0

We get rid of parentheses

r^2+4r-33/4=0

We multiply all the terms by the denominator


r^2*4+4r*4-33=0

Wy multiply elements

4r^2+16r-33=0

a = 4; b = 16; c = -33;
Δ = b2-4ac
Δ = 162-4·4·(-33)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-28}{2*4}=\frac{-44}{8}
=-5+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+28}{2*4}=\frac{12}{8}
=1+1/2
$