4r^2+29r+45=0

Solution for 4r^2+29r+45=0 equation:

4r^2+29r+45=0

a = 4; b = 29; c = +45;
Δ = b2-4ac
Δ = 292-4·4·45
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-11}{2*4}=\frac{-40}{8}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*4}=\frac{-18}{8}
=-2+1/4
$