(r+7)(4r+1)=0

Solution for (r+7)(4r+1)=0 equation:

(r+7)(4r+1)=0

We multiply parentheses ..

(+4r^2+r+28r+7)=0

We get rid of parentheses

4r^2+r+28r+7=0

We add all the numbers together, and all the variables

4r^2+29r+7=0

a = 4; b = 29; c = +7;
Δ = b2-4ac
Δ = 292-4·4·7
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*4}=\frac{-56}{8}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*4}=\frac{-2}{8}
=-1/4
$