If it's not what You are looking for type in the equation solver your own equation and let us solve it.
3x^2+39x+12=0
a = 3; b = 39; c = +12;
Δ = b2-4ac
Δ = 392-4·3·12
Δ = 1377
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1377}=\sqrt{81*17}=\sqrt{81}*\sqrt{17}=9\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-9\sqrt{17}}{2*3}=\frac{-39-9\sqrt{17}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+9\sqrt{17}}{2*3}=\frac{-39+9\sqrt{17}}{6} $
| 5n+5=-10n+20 | | 7q-q=12 | | x^2-2.5=-1.5 | | 25g^2+14g=0 | | 5n+5=-10+20 | | 1/3(3x+9)=60 | | x^2-1.5=-1.5 | | x3-2(x+3)=-8 | | 4x+4+x=3x-6+2 | | 33s^2-50s=0 | | -t-4=8=7t | | 4n^2+5n-9=0 | | 9-2x=35x= | | 2x-1+x/2=1+4x+5 | | -1-1/3w=1 | | j^2-47j=0 | | (625/256)^(5x-4)=1 | | 25x-17=-64 | | 6(5t+1)-12=t-(t+5) | | (4y-3)-3×=11 | | x8-12=-10 | | -1/2y-8=23 | | 9m+14m=69 | | j^2+9j+14=0 | | 3y-6-5y=-2y-6 | | 4(p-2)=3p-7 | | 3v+6-v=2v-2+5 | | 23y^2+47+2=0 | | (16b-3)-5(3b+4)=-4 | | 6x+29x-6=7(5x+9) | | 8x=−14 | | m^2+22m+21=0 |