0.1(t+0.5)0.2*t=0.3(t-0.4)

Solution for 0.1(t+0.5)0.2*t=0.3(t-0.4) equation:

0.1(t+0.5)0.2t=0.3(t-0.4)

We move all terms to the left:

0.1(t+0.5)0.2t-(0.3(t-0.4))=0

We multiply parentheses

0t^2+0t-(0.3(t-0.4))=0


We calculate terms in parentheses: -(0.3(t-0.4)), so:

0.3(t-0.4)

We multiply parentheses

0.3t-0.12

Back to the equation:

-(0.3t-0.12)


We add all the numbers together, and all the variables

t^2+t-(0.3t-0.12)=0

We get rid of parentheses

t^2+t-0.3t+0.12=0

We add all the numbers together, and all the variables

t^2+0.7t+0.12=0

a = 1; b = 0.7; c = +0.12;
Δ = b2-4ac
Δ = 0.72-4·1·0.12
Δ = 0.01
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{-(0.7)-\sqrt{0.01}}{2*1}=\frac{-0.7-\sqrt{0.01}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.7)+\sqrt{0.01}}{2*1}=\frac{-0.7+\sqrt{0.01}}{2}
$