If it's not what You are looking for type in the equation solver your own equation and let us solve it.
7b^2+29b+4=0
a = 7; b = 29; c = +4;
Δ = b2-4ac
Δ = 292-4·7·4
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*7}=\frac{-56}{14} =-4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*7}=\frac{-2}{14} =-1/7 $
| -138=2+10x | | -7=r/16-8 | | x^2−18x+81=64 | | 9b^2=36 | | -9+a/2=-19 | | 1.6(6-x)=10 | | 43=4n-9 | | 5f-8=11 | | (4.9x^2)+9.8x-39.2=0 | | 4=3+x/16 | | -6=-4+p/2 | | 1425=(x+35)17 | | 3x=44/3 | | 6+b/10=5 | | /5(x-3)=15 | | -13=-6+x/2 | | -10=-10x-10 | | X^3-8x^2+4x-12=0 | | b/8+2=4 | | 3p+4=-35 | | -1+3b=-52 | | 7(a+6)=56 | | 2+m/15=3 | | -3=-1+v/7 | | (50x-15)=(42x+1) | | 5(3y+2=10 | | 2x±8=16 | | 1y-y=0y | | 19a-12-4a= | | (6x+15)=(8x-23) | | 4(v+3)=40 | | -1+n=-18 |