If it's not what You are looking for type in the equation solver your own equation and let us solve it.
8z^2+23z+3=0
a = 8; b = 23; c = +3;
Δ = b2-4ac
Δ = 232-4·8·3
Δ = 433
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{433}}{2*8}=\frac{-23-\sqrt{433}}{16} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{433}}{2*8}=\frac{-23+\sqrt{433}}{16} $
| 10+13y=11y+2 | | 3-(2x-8)=2x-5 | | 14+8a=3a+12 | | 13a+5=4a+6 | | (2x-1)×(4x-3)=0 | | 7z+2=6+9z | | 2x3-3X+10=0 | | 4x=3+-25 | | -2p=-5/2 | | Y2+3y=5 | | 22x=3x | | a+6+6=11 | | x2+13+42=0 | | 3x-12x=15 | | 5-y/2=3y-1 | | A=-1+a | | 2x+8-7=12 | | 19=-9y-5×6y | | 1.2(y-2)=2y-1.2 | | 1.2y-2=2y-1.2 | | (z+9)^2=17 | | 5(3,2x+4,3)=69,5 | | 5e+14=0 | | 5(7d+2)=-4d | | -3z-1=2(z+7)$ | | 6/5+5/4x=2/3x | | 3x-(12-4x)=2(x-15) | | 4-2(x+3)=x-5+6x | | 24-d=8 | | x÷7=6 | | 3(2b-5)=17 | | 7(k-11)=2(11+k) |