If it's not what You are looking for type in the equation solver your own equation and let us solve it.
7z^2+29z-30=0
a = 7; b = 29; c = -30;
Δ = b2-4ac
Δ = 292-4·7·(-30)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1681}=41$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*7}=\frac{-70}{14} =-5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*7}=\frac{12}{14} =6/7 $
| 4c–11=5c+7 | | 5x-10+4x=71 | | 6+a=-17 | | 5t2-2t-3=0 | | 8+4x=1+2x | | 4x-5/1-3x=x-3/x-5 | | 16-2t=3|2t+9 | | 25=9x=19 | | K+1-1k=-2k | | 7b-7-8b=15 | | 14.7x-5=10x-x-15 | | 12.20-9x=x+10x | | 1/10y+2=-6 | | 11.x+1+4x=11+12x-3 | | 14x-50=16x-49-1-2x | | 10.1+8x+9=-20+2x | | 30=2(8+x)-8(6x+4) | | 2+6n=22 | | 3k+7k-5=45 | | 5/6s=420 | | (x+2)x5=20 | | -16+x/13=-4 | | 4=b+14 | | -2(y+3)+6=6+5y | | 0=-1x^2-2x+7 | | 7.x-2=10-5x | | 1/2x+2=3/4x+2+1/2 | | 3(5+x)=20+ | | 14+5=-5x-6(-3x=15)+5 | | 6x=26.69 | | -7y+4=-31 | | -15=-5x+3(x-3 |