(1/8)(8t-7)=t+(7/16)

Solution for (1/8)(8t-7)=t+(7/16) equation:

(1/8)(8t-7)=t+(7/16)

We move all terms to the left:

(1/8)(8t-7)-(t+(7/16))=0


Domain of the equation: 8)(8t-7)!=0

t∈R


We add all the numbers together, and all the variables

(+1/8)(8t-7)-(t+(+7/16))=0

We multiply parentheses ..

(+8t^2+1/8*-7)-(t+(+7/16))=0

We calculate fractions


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

We add all the numbers together, and all the variables

8t^2/()+(-1t)/()=0

We multiply all the terms by the denominator


8t^2+(-1t)=0

We get rid of parentheses

8t^2-1t=0

a = 8; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·8·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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