0=4(4t^2-31t+21)

Solution for 0=4(4t^2-31t+21) equation:

0=4(4t^2-31t+21)

We move all terms to the left:

0-(4(4t^2-31t+21))=0

We add all the numbers together, and all the variables

-(4(4t^2-31t+21))=0


We calculate terms in parentheses: -(4(4t^2-31t+21)), so:

4(4t^2-31t+21)

We multiply parentheses

16t^2-124t+84

Back to the equation:

-(16t^2-124t+84)


We get rid of parentheses

-16t^2+124t-84=0

a = -16; b = 124; c = -84;
Δ = b2-4ac
Δ = 1242-4·(-16)·(-84)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(124)-100}{2*-16}=\frac{-224}{-32}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(124)+100}{2*-16}=\frac{-24}{-32}
=3/4
$