(0.7t-0.3)(0.11t+1)=0

Solution for (0.7t-0.3)(0.11t+1)=0 equation:

(0.7t-0.3)(0.11t+1)=0

We multiply parentheses ..

(+0t^2+0t-0t-0.3)=0

We get rid of parentheses

0t^2+0t-0t-0.3=0

We add all the numbers together, and all the variables

t^2-0.3=0

a = 1; b = 0; c = -0.3;
Δ = b2-4ac
Δ = 02-4·1·(-0.3)
Δ = 1.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{1.2}}{2*1}=\frac{0-\sqrt{1.2}}{2}
=-\frac{\sqrt{}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.2}}{2*1}=\frac{0+\sqrt{1.2}}{2}
=\frac{\sqrt{}}{2}
$