3/4t=17t+12

Solution for 3/4t=17t+12 equation:

3/4t=17t+12

We move all terms to the left:

3/4t-(17t+12)=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R


We get rid of parentheses

3/4t-17t-12=0

We multiply all the terms by the denominator


-17t*4t-12*4t+3=0

Wy multiply elements

-68t^2-48t+3=0

a = -68; b = -48; c = +3;
Δ = b2-4ac
Δ = -482-4·(-68)·3
Δ = 3120
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:

$\sqrt{\Delta}=\sqrt{3120}=\sqrt{16*195}=\sqrt{16}*\sqrt{195}=4\sqrt{195}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{195}}{2*-68}=\frac{48-4\sqrt{195}}{-136}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{195}}{2*-68}=\frac{48+4\sqrt{195}}{-136}
$