If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5k^2+12k-108=0
a = 5; b = 12; c = -108;
Δ = b2-4ac
Δ = 122-4·5·(-108)
Δ = 2304
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{2304}=48$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-48}{2*5}=\frac{-60}{10} =-6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+48}{2*5}=\frac{36}{10} =3+3/5 $
| 5t^2+23t+10=0 | | 9/4x+7=½-x | | 2(-x-1)+5=x+10 | | 24x+8-4x+1=9 | | 3(x+20)+44=180 | | 2n2+n-55=0 | | X-4=3x*8 | | 5.2-3.05x=1+6.15x | | 3r2-11r+8=0 | | 4(x-3)+36=64-4(x+2) | | n2+7n+12=0 | | -8=1.3-2.1m | | -9y²=-144 | | x-7-1=-10+4 | | 1/9t^2-1=0 | | 6n2-7n-49=0 | | q+2=8 | | 3(x+2)-2x=15 | | 2(u+4)=-2u+44 | | 14+w-11=19 | | (1/2)f-2=(1/6)f | | -11=-9-x3 | | 2x-8=4(-2-3x) | | -16+x/4=32 | | X2+4y2=8 | | (4x+8)=(7x+2/4) | | 6a2+2a-48=0 | | 4y+12=-6y | | 1440=1/2x(40x+120x) | | 6x+15=8x+1 | | x-7/3=3/4 | | (5y-2)÷4=3 |