If it's not what You are looking for type in the equation solver your own equation and let us solve it.
15z^2+16z+4=0
a = 15; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·15·4
Δ = 16
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{16}=4$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*15}=\frac{-20}{30} =-2/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*15}=\frac{-12}{30} =-2/5 $
| 2(t-4)+4=2(2t-4) | | 5g^2+25g-180=0 | | F(x)=3.2x-0.6 | | 37.1f=7.0 | | 3(x+2)=(2x-5) | | X²+4x+480=0 | | 28x+8=35x+280 | | 3x+×=40 | | f(1)=10-30/1+5 | | (5+x)/5=9/8 | | 1/2(u)=25 | | 2m-10=5m-80 | | 12x+72+6x+36=11 | | 3y-2/4-2y+3/3=4 | | 54=3z-10 | | x+5/4+9=7 | | 5m-8=4m+3 | | 3y+4+3Y=34 | | (3x)+55=133 | | 4/9x-1/3=0 | | (3x)=55 | | 4+2b+12=0 | | 9y-4=5y=8 | | 2(3y-3)=12 | | 2n-19=-37 | | 4(3/10)-(22/5x+51/2)=1/2(-33/5x+11/5) | | 5b-11=60 | | 2n+21=15 | | 12-4/5x=4 | | |2c-1|=-11 | | 10/(10+x)=0.625 | | y=3600•(1.04)^25 |