t2-2t+(3/4)=0

Solution for t2-2t+(3/4)=0 equation:

t2-2t+(3/4)=0

We add all the numbers together, and all the variables

t2-2t+(+3/4)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

t^2-2t+3/4=0

We multiply all the terms by the denominator


t^2*4-2t*4+3=0

Wy multiply elements

4t^2-8t+3=0

a = 4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·4·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*4}=\frac{4}{8}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8}
=1+1/2
$