(4t+9)(t+4)=0

Solution for (4t+9)(t+4)=0 equation:

(4t+9)(t+4)=0

We multiply parentheses ..

(+4t^2+16t+9t+36)=0

We get rid of parentheses

4t^2+16t+9t+36=0

We add all the numbers together, and all the variables

4t^2+25t+36=0

a = 4; b = 25; c = +36;
Δ = b2-4ac
Δ = 252-4·4·36
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*4}=\frac{-32}{8}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*4}=\frac{-18}{8}
=-2+1/4
$