If it's not what You are looking for type in the equation solver your own equation and let us solve it.
6j^2+23j+7=0
a = 6; b = 23; c = +7;
Δ = b2-4ac
Δ = 232-4·6·7
Δ = 361
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{361}=19$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*6}=\frac{-42}{12} =-3+1/2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*6}=\frac{-4}{12} =-1/3 $
| 7.4^(x+7)+3=8 | | 4u/7=16 | | 33/4+11/3=x | | 16x-9x-2x=27 | | 5x+17=4(3x−1 | | 8(x+1)-6(x+2)=20 | | 5x+17=2x-3 | | 24/x-1=8 | | 3/y=7y | | 12x-3(x+6)=11x-10 | | 49t^2=4 | | -62/9=5x-2x-8/9 | | 6(3n+1)=7(9n+4)+1 | | 2x-25+4x-11+6+x+3x=360 | | 4j-20=10j-20 | | 4(3x+2)=-3(4x-3) | | 0.02+0.03x=0.77 | | 5(x+6)+6=31 | | 3x-32=12+x | | 5x-(4x÷5)=5 | | 6+3x-5/7x=-38/7 | | g^2+11g–12=0 | | 54=6-8d=d | | 9-x-7x=7 | | 130=4x-5 | | d2+18d=0 | | 1/2+2x=11/14 | | 35+6p=-5(-4p-7) | | 8x+21=90 | | 130=(4x-5) | | -5(x-5)=-3x | | 2-2x3-3=X |