(5/2)a-(a+4)=a+10

Solution for (5/2)a-(a+4)=a+10 equation:

(5/2)a-(a+4)=a+10

We move all terms to the left:

(5/2)a-(a+4)-(a+10)=0


Domain of the equation: 2)a!=0

a!=0/1

a!=0

a∈R


We add all the numbers together, and all the variables

(+5/2)a-(a+4)-(a+10)=0

We multiply parentheses

5a^2-(a+4)-(a+10)=0

We get rid of parentheses

5a^2-a-a-4-10=0

We add all the numbers together, and all the variables

5a^2-2a-14=0

a = 5; b = -2; c = -14;
Δ = b2-4ac
Δ = -22-4·5·(-14)
Δ = 284
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{71}}{2*5}=\frac{2-2\sqrt{71}}{10}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{71}}{2*5}=\frac{2+2\sqrt{71}}{10}
$