(3/4)x+(1/4)x=19

Solution for (3/4)x+(1/4)x=19 equation:

(3/4)x+(1/4)x=19

We move all terms to the left:

(3/4)x+(1/4)x-(19)=0


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)x+(+1/4)x-19=0

We multiply parentheses

3x^2+x^2-19=0

We add all the numbers together, and all the variables

4x^2-19=0

a = 4; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·4·(-19)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{19}}{2*4}=\frac{0-4\sqrt{19}}{8}
=-\frac{4\sqrt{19}}{8}
=-\frac{\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{19}}{2*4}=\frac{0+4\sqrt{19}}{8}
=\frac{4\sqrt{19}}{8}
=\frac{\sqrt{19}}{2}
$