(1/8)t+(1/6)t=7

Solution for (1/8)t+(1/6)t=7 equation:

(1/8)t+(1/6)t=7

We move all terms to the left:

(1/8)t+(1/6)t-(7)=0


Domain of the equation: 8)t!=0

t!=0/1

t!=0

t∈R



Domain of the equation: 6)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+1/8)t+(+1/6)t-7=0

We multiply parentheses

t^2+t^2-7=0

We add all the numbers together, and all the variables

2t^2-7=0

a = 2; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·2·(-7)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*2}=\frac{0-2\sqrt{14}}{4}
=-\frac{2\sqrt{14}}{4}
=-\frac{\sqrt{14}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*2}=\frac{0+2\sqrt{14}}{4}
=\frac{2\sqrt{14}}{4}
=\frac{\sqrt{14}}{2}
$