If it's not what You are looking for type in the equation solver your own equation and let us solve it.
X+3/4X=41
We move all terms to the left:
X+3/4X-(41)=0
Domain of the equation: 4X!=0We multiply all the terms by the denominator
X!=0/4
X!=0
X∈R
X*4X-41*4X+3=0
Wy multiply elements
4X^2-164X+3=0
a = 4; b = -164; c = +3;
Δ = b2-4ac
Δ = -1642-4·4·3
Δ = 26848
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{26848}=\sqrt{16*1678}=\sqrt{16}*\sqrt{1678}=4\sqrt{1678}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-164)-4\sqrt{1678}}{2*4}=\frac{164-4\sqrt{1678}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-164)+4\sqrt{1678}}{2*4}=\frac{164+4\sqrt{1678}}{8} $
| 2.7.b=54 | | 27+3(5x-7)=47 | | -5x-9=-0 | | 3.q-145.5=5q | | w-4.48=3.31 | | 6(z+3)–9=27 | | 3x-1=-37 | | x+3.1=9.83 | | -2b^2+5b=0 | | 4y-1=3(-5-7) | | (5k+3)^2=27 | | (5k+3)^3=27 | | f(-4)=-2-5 | | -13=12(x/4) | | 5y-8+y+52=6y+50-2y | | 51+(8x+1)=90 | | 3q^2-5q-6=3q^2 | | -13=12x/4 | | b/7+2=9 | | 2x+6+3x-18+3x-18+3x-9+3x-9=360 | | a/10+5=3 | | 14x^2-2x=3-21x | | 11j−10j=17 | | 3q^2-14q+12=4 | | 4√3x2+5x2-2√3=0 | | 4m+(39/45)=4 | | (x+84)+(2x)=180 | | 14=3x-0.1 | | 50-66y=90 | | 0.2m=-3(m+330) | | -65-7x=264 | | 4−-2k=-10 |