(3/4)x+(2/8)x=48/8

Solution for (3/4)x+(2/8)x=48/8 equation:

(3/4)x+(2/8)x=48/8

We move all terms to the left:

(3/4)x+(2/8)x-(48/8)=0


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 8)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)x+(+2/8)x-6=0

We multiply parentheses

3x^2+2x^2-6=0

We add all the numbers together, and all the variables

5x^2-6=0

a = 5; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·5·(-6)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*5}=\frac{0-2\sqrt{30}}{10}
=-\frac{2\sqrt{30}}{10}
=-\frac{\sqrt{30}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*5}=\frac{0+2\sqrt{30}}{10}
=\frac{2\sqrt{30}}{10}
=\frac{\sqrt{30}}{5}
$