If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(cos(x))^2
4-(2*(3*x-(1/2))) = 5*x-2
0.91;0.723
5*x^2-(2*x)+0*x^3-3 = 0
((6*a^-4)/3)*a^-3 = 0
| 7b+6b-10= | | -7+y=-1 | | 4-2(3x-1/2)=5x-2 | | -3x+7=115+6x | | 183=39+0.30x | | 5x=3-13 | | -8+r=-9 | | F(x)=x^4-4 | | a=0.5b-4 | | 1-7x=5x+145 | | 6a^-4/3a^-3 | | x=14+0.875y | | (0.075)(-5.5)= | | 3(-4x+8)=0 | | 0.13(y-3)+0.21y=0.04y-0.30 | | 6g+24=6g+12 | | -8x-3=2x-48 | | (-5)x(-7)= | | (6y+6)-(5+5y)=4 | | (5/6)=.75x | | z-3.2=1.6 | | 5b+2b-3=11 | | 3x-36=x+12 | | 5x-9x+62=3x+27 | | 15=10-1/2k | | 5(3x+9)=15x-3 | | f(x)=log(2x+3) | | 3x+3=216 | | 80=40+.35x | | 6(3x+2)-12=108 | | 31+4t-1=40 | | 42=n+27 |