If it's not what You are looking for type in the equation solver your own equation and let us solve it.
3y^2+11y+-4=0
We add all the numbers together, and all the variables
3y^2+11y=0
a = 3; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·3·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*3}=\frac{-22}{6} =-3+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*3}=\frac{0}{6} =0 $
| 6(t-4)=-6 | | 9(x-2)/6=6(x+1)/10 | | x+7-8x/3=17/6-5x/8x+7-8x/3=17/6-5x/8 | | (p-3)(p+2)=14 | | 5(4s+3)=115 | | x(1-12/x)2=11/25 | | 4x^2+12x=154 | | 12u=7u+15 | | 41/2*x=11/2 | | 3(x-1)/2=3(x+7)/10 | | 4½*x=1½ | | 3(x-1)/2=3((x+7)/10 | | x*1/2=1/3 | | 1/2*x=3/8 | | ⅓*x=⅔ | | 4(y+8)+y=3(y+2)+7 | | 9(y+8)+y=4(y+3)-2 | | 4(y+7)+y=3(y+3)+4 | | K^2-18k=13 | | K^2-18k=0 | | -3(t-2)+7t=6t-2 | | -3(t-1)+6t=8t-8 | | 11p+19=140 | | -3(t-2)+9t=3t-7 | | 7p-1=90 | | 9p-15=66 | | -16t+12t+40=0 | | 7(8-y)/5=-y | | 3(x-1)=2x-1/2 | | 11y+6=9 | | 5(1-y)/2=y | | E(4)=20h |