(70/8)=2t^2-t

Solution for (70/8)=2t^2-t equation:

(70/8)=2t^2-t

We move all terms to the left:

(70/8)-(2t^2-t)=0

We add all the numbers together, and all the variables

-(2t^2-t)+(+70/8)=0

We get rid of parentheses

-2t^2+t+70/8=0

We multiply all the terms by the denominator


-2t^2*8+t*8+70=0

Wy multiply elements

-16t^2+8t+70=0

a = -16; b = 8; c = +70;
Δ = b2-4ac
Δ = 82-4·(-16)·70
Δ = 4544
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:

$\sqrt{\Delta}=\sqrt{4544}=\sqrt{64*71}=\sqrt{64}*\sqrt{71}=8\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{71}}{2*-16}=\frac{-8-8\sqrt{71}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{71}}{2*-16}=\frac{-8+8\sqrt{71}}{-32}
$