8t^2+4t+7/8=2

Solution for 8t^2+4t+7/8=2 equation:

8t^2+4t+7/8=2

We move all terms to the left:

8t^2+4t+7/8-(2)=0

determiningTheFunctionDomain
8t^2+4t-2+7/8=0

We multiply all the terms by the denominator


8t^2*8+4t*8+7-2*8=0

We add all the numbers together, and all the variables

8t^2*8+4t*8-9=0

Wy multiply elements

64t^2+32t-9=0

a = 64; b = 32; c = -9;
Δ = b2-4ac
Δ = 322-4·64·(-9)
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{13}}{2*64}=\frac{-32-16\sqrt{13}}{128}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{13}}{2*64}=\frac{-32+16\sqrt{13}}{128}
$