If it's not what You are looking for type in the equation solver your own equation and let us solve it.
1/2q+2(q+5)=-4(q(1)+1
We move all terms to the left:
1/2q+2(q+5)-(-4(q(1)+1)=0
Domain of the equation: 2q!=0We add all the numbers together, and all the variables
q!=0/2
q!=0
q∈R
1/2q+2(q+5)-(-4(q+1)=0
We multiply parentheses
1/2q+2q-(-4(q+1)+10=0
We multiply all the terms by the denominator
2q*2q-((-4(q+1)+10)*2q+1=0
Wy multiply elements
4q^2-((-4(q+1)+10)*2q+1=0
| (3/2)x+-4=13 | | Y=-20+2x | | X+.06x=5 | | 4x-(3x+1)=-11 | | (3+5w+3w²)^6=64 | | x-5x=-36 | | a+4a+5a-12=180 | | 3.6x-4.6=2.6 | | 7x+6x-40=60-7x | | 5(3-3y/2)+4y=13.25 | | 4=x+9÷13 | | 20=30x+30 | | -4x8=-2x+12 | | y=2(4)^-3 | | y=2(4)^-2 | | y=20000(0.95)^10 | | y=20000(0.95)^4 | | y=20000(0.95)^3 | | y=20000(0.95)^2 | | y=20000(0.95)^1 | | y=12841(1.01)^20 | | -4(y+4)=2y-8+6(2y-1) | | y=12841(1.01)^4 | | y=12841(1.01)^3 | | y=12841(1.01)^2 | | 25c-9=0 | | 2=10x-21 | | y=12841(1.04) | | y=12841(1.03) | | y=12841(1.02) | | y=12841(2.01) | | 7(u-1)=7u+1-2(-9u-2) |