(21/2)x+(31/2)x=18

Solution for (21/2)x+(31/2)x=18 equation:

(21/2)x+(31/2)x=18

We move all terms to the left:

(21/2)x+(31/2)x-(18)=0


Domain of the equation: 2)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+21/2)x+(+31/2)x-18=0

We multiply parentheses

21x^2+31x^2-18=0

We add all the numbers together, and all the variables

52x^2-18=0

a = 52; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·52·(-18)
Δ = 3744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3744}=\sqrt{144*26}=\sqrt{144}*\sqrt{26}=12\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{26}}{2*52}=\frac{0-12\sqrt{26}}{104}
=-\frac{12\sqrt{26}}{104}
=-\frac{3\sqrt{26}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{26}}{2*52}=\frac{0+12\sqrt{26}}{104}
=\frac{12\sqrt{26}}{104}
=\frac{3\sqrt{26}}{26}
$