(t-3)2=(4-3t)/4

Solution for (t-3)2=(4-3t)/4 equation:

(t-3)2=(4-3t)/4

We move all terms to the left:

(t-3)2-((4-3t)/4)=0

We add all the numbers together, and all the variables

(t-3)2-((-3t+4)/4)=0

We multiply parentheses

2t-((-3t+4)/4)-6=0

We multiply all the terms by the denominator


2t*4)-((-3t+4)-6*4)=0

We add all the numbers together, and all the variables

2t*4)-((-3t+4)=0

Wy multiply elements

8t^2=0

a = 8; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·8·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{0}{16}=0$