If it's not what You are looking for type in the equation solver your own equation and let us solve it.
-12x^2+260x=0
a = -12; b = 260; c = 0;
Δ = b2-4ac
Δ = 2602-4·(-12)·0
Δ = 67600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{67600}=260$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(260)-260}{2*-12}=\frac{-520}{-24} =21+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(260)+260}{2*-12}=\frac{0}{-24} =0 $
| y=09*0.266-1E+07 | | 28-7x/12=x/6 | | 3/7m-1/2=4/5 | | 5.1q-4.3-6.2q=-2.8-1.1q-1.5 | | 5.1q-4.3-6.2q=-2.8-1.1q-15 | | 11(t-3)+3t=7(2t+2)-10 | | -(3x+2)-(2x-9)=13 | | -5+k=1 | | 6(d-8)=3(14=d) | | 3.7x-3=2.7x | | 4x-500x^-1/2=0 | | 2x+10=2x+8 | | 1/2(18x+8)-2=-3/4(24x-32) | | 11y-12=3y+12 | | 2h/7=4 | | 5h+15=45 | | 1/2(16x+8)-11=-1/3(18x-18) | | x(12-x)^3=0 | | F(t)=√t+2 | | F(t)=2t³-5t | | 9(x+4)=5x-(4-x) | | F(t)=t^2-4t-7 | | 10x-(5x-4)=24 | | H(t)=58t-0.82t^2 | | Y(t)=40t-16t^2 | | -1/8m+1-13/8m=5 | | 5x=1,255 | | F(t)=0.2t^2+2t | | 1.07^x=3 | | 15+u-2=9^2 | | (5-2/3x)=1 | | 1.6=0.9*x |