(2/5)t-3=1

Solution for (2/5)t-3=1 equation:

(2/5)t-3=1

We move all terms to the left:

(2/5)t-3-(1)=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

(+2/5)t-3-1=0

We add all the numbers together, and all the variables

(+2/5)t-4=0

We multiply parentheses

2t^2-4=0

a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4}
=-\frac{4\sqrt{2}}{4}
=-\sqrt{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4}
=\frac{4\sqrt{2}}{4}
=\sqrt{2}
$