(2t+7)(4-3t)=0

Solution for (2t+7)(4-3t)=0 equation:

(2t+7)(4-3t)=0

We add all the numbers together, and all the variables

(2t+7)(-3t+4)=0

We multiply parentheses ..

(-6t^2+8t-21t+28)=0

We get rid of parentheses

-6t^2+8t-21t+28=0

We add all the numbers together, and all the variables

-6t^2-13t+28=0

a = -6; b = -13; c = +28;
Δ = b2-4ac
Δ = -132-4·(-6)·28
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-29}{2*-6}=\frac{-16}{-12}
=1+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+29}{2*-6}=\frac{42}{-12}
=-3+1/2
$