(4t+7)(t+6)=0

Solution for (4t+7)(t+6)=0 equation:

(4t+7)(t+6)=0

We multiply parentheses ..

(+4t^2+24t+7t+42)=0

We get rid of parentheses

4t^2+24t+7t+42=0

We add all the numbers together, and all the variables

4t^2+31t+42=0

a = 4; b = 31; c = +42;
Δ = b2-4ac
Δ = 312-4·4·42
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-17}{2*4}=\frac{-48}{8}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+17}{2*4}=\frac{-14}{8}
=-1+3/4
$