If it's not what You are looking for type in the equation solver your own equation and let us solve it.
28j^2+12j-1=0
a = 28; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·28·(-1)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*28}=\frac{-28}{56} =-1/2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*28}=\frac{4}{56} =1/14 $
| 5v-5=30 | | g~-5+3=-13 | | 9(g+43)-68=-59 | | 8x+3=19= | | 50+(x+30)+3x=180 | | -5-6k=-6-6k | | c–5.3=–6.4 | | 3(r-64)-29=4 | | 52=18t-4.6t^2 | | 1/2x=6/3 | | 76=8(m+8)+4 | | (X1/3)3=c | | 2(4e-3)-8-2e=4 | | -5=1/2(x+4)-3/4x | | 89-c=2 | | –14+t=26 | | 3x=8x-5x | | y+3/4(1.15)-2=180 | | 26-(4x-11)=-5(x+7 | | 3e+6e-7e=16 | | 6(7x=7)=7(6x=6) | | 5(x+2)+2(x-2)=48 | | 16=d-12~14 | | 3(j+19)-17=-14 | | 5x+7=-2×+21 | | 3y-3=16 | | 2w-54+w=21 | | 7x-42=3x-2 | | -35=-7/5x | | 6x-5-2x=x-5 | | -1/2x=-1/2x+4 | | 9x-11-x+4=2x+38+x |