16-2(t)=3/2(t)+9

Solution for 16-2(t)=3/2(t)+9 equation:

16-2(t)=3/2(t)+9

We move all terms to the left:

16-2(t)-(3/2(t)+9)=0


Domain of the equation: 2t+9)!=0

t∈R


We get rid of parentheses

-2t-3/2t-9+16=0

We multiply all the terms by the denominator


-2t*2t-9*2t+16*2t-3=0

Wy multiply elements

-4t^2-18t+32t-3=0

We add all the numbers together, and all the variables

-4t^2+14t-3=0

a = -4; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·(-4)·(-3)
Δ = 148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{37}}{2*-4}=\frac{-14-2\sqrt{37}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{37}}{2*-4}=\frac{-14+2\sqrt{37}}{-8}
$