4+-4/4+-7=2t+5/2t-5

Solution for 4+-4/4+-7=2t+5/2t-5 equation:

4+-4/4+-7=2t+5/2t-5

We move all terms to the left:

4+-4/4+-7-(2t+5/2t-5)=0


Domain of the equation: 2t-5)!=0

t∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-2t-5/2t+5=0

We multiply all the terms by the denominator


-2t*2t+5*2t-5=0

Wy multiply elements

-4t^2+10t-5=0

a = -4; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·(-4)·(-5)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*-4}=\frac{-10-2\sqrt{5}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*-4}=\frac{-10+2\sqrt{5}}{-8}
$