If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(3q^2)-10q=0
a = 3; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·3·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*3}=\frac{0}{6} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*3}=\frac{20}{6} =3+1/3 $
| 3q^23-10q=0 | | 83=3u | | P(x)=1(x)+5(x)-3 | | -5(c-10)+3c=25 | | m^2+37m=0 | | 2x+21+2x+21=x+21 | | 5x+4-60=75+7x | | x+0.05*x=14700 | | 180=90+(3x+2) | | 10g−9g−g+3g=12 | | 9x-1=2x-1 | | 12+2(x+3)=3x-6 | | 9(5+n)=81 | | x(.6)=450 | | 3m+87-5m=-150 | | 9(5+n=81 | | 2^x=3000 | | 325=a^2 | | 0.20=x^2/25-x | | x+0.05*x=13000 | | (p^2)+23p=0 | | 20=t/5 | | (4x+7)(2x+5)=0 | | 29d^2-343d+84=0 | | 3y-28+11+11=180 | | (c-3)=(c²-2c+3) | | 300x-3=120+2 | | 10=|3x+7| | | 3y-28+11=180 | | 3x-22+77+77=180 | | (8j^2)-21j=0 | | 6x-9=12-7 |