r^2+(3/4)r-5=0

Solution for r^2+(3/4)r-5=0 equation:

r^2+(3/4)r-5=0


Domain of the equation: 4)r!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

r^2+(+3/4)r-5=0

We multiply parentheses

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

We add all the numbers together, and all the variables

4r^2-5=0

a = 4; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·4·(-5)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*4}=\frac{0-4\sqrt{5}}{8}
=-\frac{4\sqrt{5}}{8}
=-\frac{\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*4}=\frac{0+4\sqrt{5}}{8}
=\frac{4\sqrt{5}}{8}
=\frac{\sqrt{5}}{2}
$