(7/8)+(1/5)t=2

Solution for (7/8)+(1/5)t=2 equation:

(7/8)+(1/5)t=2

We move all terms to the left:

(7/8)+(1/5)t-(2)=0


Domain of the equation: 5)t!=0

t!=0/1

t!=0

t∈R


determiningTheFunctionDomain
(1/5)t-2+(7/8)=0

We add all the numbers together, and all the variables

(+1/5)t-2+(+7/8)=0

We multiply parentheses

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

We get rid of parentheses

t^2-2+7/8=0

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

t^2*8-9=0

Wy multiply elements

8t^2-9=0

a = 8; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·8·(-9)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*8}=\frac{0-12\sqrt{2}}{16}
=-\frac{12\sqrt{2}}{16}
=-\frac{3\sqrt{2}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*8}=\frac{0+12\sqrt{2}}{16}
=\frac{12\sqrt{2}}{16}
=\frac{3\sqrt{2}}{4}
$