1/4t-11+1/2t+7=16+(-10)

Solution for 1/4t-11+1/2t+7=16+(-10) equation:

1/4t-11+1/2t+7=16+(-10)

We move all terms to the left:

1/4t-11+1/2t+7-(16+(-10))=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R



Domain of the equation: 2t!=0

t!=0/2

t!=0

t∈R


We add all the numbers together, and all the variables

1/4t+1/2t-10=0

We calculate fractions


2t/8t^2+4t/8t^2-10=0

We multiply all the terms by the denominator


2t+4t-10*8t^2=0

We add all the numbers together, and all the variables

6t-10*8t^2=0

Wy multiply elements

-80t^2+6t=0

a = -80; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-80)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-80}=\frac{-12}{-160}
=3/40
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-80}=\frac{0}{-160}
=0
$