(4/5)t-2=10

Solution for (4/5)t-2=10 equation:

(4/5)t-2=10

We move all terms to the left:

(4/5)t-2-(10)=0


Domain of the equation: 5)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+4/5)t-2-10=0

We add all the numbers together, and all the variables

(+4/5)t-12=0

We multiply parentheses

4t^2-12=0

a = 4; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4·(-12)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*4}=\frac{0-8\sqrt{3}}{8}
=-\frac{8\sqrt{3}}{8}
=-\sqrt{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*4}=\frac{0+8\sqrt{3}}{8}
=\frac{8\sqrt{3}}{8}
=\sqrt{3}
$