-3(2t-5)/2t=5t+4

Solution for -3(2t-5)/2t=5t+4 equation:

-3(2t-5)/2t=5t+4

We move all terms to the left:

-3(2t-5)/2t-(5t+4)=0


Domain of the equation: 2t!=0

t!=0/2

t!=0

t∈R


We get rid of parentheses

-3(2t-5)/2t-5t-4=0

We multiply all the terms by the denominator


-3(2t-5)-5t*2t-4*2t=0

We multiply parentheses

-6t-5t*2t-4*2t+15=0

Wy multiply elements

-10t^2-6t-8t+15=0

We add all the numbers together, and all the variables

-10t^2-14t+15=0

a = -10; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·(-10)·15
Δ = 796
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:

$\sqrt{\Delta}=\sqrt{796}=\sqrt{4*199}=\sqrt{4}*\sqrt{199}=2\sqrt{199}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{199}}{2*-10}=\frac{14-2\sqrt{199}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{199}}{2*-10}=\frac{14+2\sqrt{199}}{-20}
$