4*2^2t=15

Solution for 4*2^2t=15 equation:

4*2^2t=15

We move all terms to the left:

4*2^2t-(15)=0

Wy multiply elements

8t^2-15=0

a = 8; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·8·(-15)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*8}=\frac{0-4\sqrt{30}}{16}
=-\frac{4\sqrt{30}}{16}
=-\frac{\sqrt{30}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*8}=\frac{0+4\sqrt{30}}{16}
=\frac{4\sqrt{30}}{16}
=\frac{\sqrt{30}}{4}
$