(4t)=2(4t)^2-7

Solution for (4t)=2(4t)^2-7 equation:

(4t)=2(4t)^2-7

We move all terms to the left:

(4t)-(2(4t)^2-7)=0

We get rid of parentheses

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

a = -24; b = 4; c = +7;
Δ = b2-4ac
Δ = 42-4·(-24)·7
Δ = 688
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{43}}{2*-24}=\frac{-4-4\sqrt{43}}{-48}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{43}}{2*-24}=\frac{-4+4\sqrt{43}}{-48}
$