5/4t+3=18,t=

Solution for 5/4t+3=18,t= equation:

5/4t+3=18.t=

We move all terms to the left:

5/4t+3-(18.t)=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R


We add all the numbers together, and all the variables

5/4t-(+18.t)+3=0

We get rid of parentheses

5/4t-18.t+3=0

We multiply all the terms by the denominator


-(18.t)*4t+3*4t+5=0

We add all the numbers together, and all the variables

-(+18.t)*4t+3*4t+5=0

We multiply parentheses

-72t^2+3*4t+5=0

Wy multiply elements

-72t^2+12t+5=0

a = -72; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·(-72)·5
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{11}}{2*-72}=\frac{-12-12\sqrt{11}}{-144}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{11}}{2*-72}=\frac{-12+12\sqrt{11}}{-144}
$