If it's not what You are looking for type in the equation solver your own equation and let us solve it.
10k^2+3k-4=0
a = 10; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·10·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*10}=\frac{-16}{20} =-4/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*10}=\frac{10}{20} =1/2 $
| 11+u=2 | | 3=b2 | | 11=11+a2 | | 2x*x=30 | | (3x+19)(7x+5)+(2x+8)=180 | | –10s+2s=–40 | | −1.5| | | 19=w+20 | | H=39.2t-4.9t² | | 1/10+1/r=1/5 | | 3w–13=14 | | 1/5=1/10+1/r | | -4.9x^2-2x+100=0 | | 35=–2s+19 | | ⅓(3X+9)=2x-10 | | 9c=–18 | | C=11x+18800 | | -3/4(8n+12)=-3n+n-(-1) | | 5c=–30 | | 7+r=–3 | | 17+c=29 | | v+14=21 | | 45=–9z | | (4x)+(9x+17)=(19x-63) | | 3+5y=-26 | | 3x^2+15=21 | | 2=2(d–3) | | –t2=6 | | x=100/150/30 | | 8=x/9+5 | | 4c-5+14c=27 | | -2(24)+15y=24 |