If it's not what You are looking for type in the equation solver your own equation and let us solve it.
20x^2-40x=0
a = 20; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·20·0
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*20}=\frac{0}{40} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*20}=\frac{80}{40} =2 $
| 1w=w | | 27+2x+8+27+7x-2=180 | | 4.9t^2-8t+50=0 | | 8x2-40x=0 | | 2/3(3a+1)=5 | | 9(z+4)-4(2+1)=2(z-1)+2(z-3) | | 5x2-245=0 | | 3/5(2f-10)=9 | | 1-8r=16-5r | | 5x+27=9-3 | | 3/4(2d+6)=5 | | 15x2-90x=0 | | 2/3(3q-12)=-5 | | 11(z+1)-4(z-4)=2(z+4)+4(z+1) | | 1/2(6x+10)=-4 | | 6y-4×=3 | | 9f-4+6=35 | | 6x2+54x=0 | | 6x6=56 | | -4/3n-9=-6 | | 10-x=3x-18 | | 10=4+3{t+2} | | 1x/4-5=1 | | 1/4x+8=3/8+6 | | 9(z+4)−3(z−4)=2(z−4)+3(z−1) | | -3+6/5r=-15 | | 9y+7y=17 | | 8x2-72x=0 | | 4(2x-7)-3=4(x-4)+9 | | 6/7=9/a | | (2x-4)+(4x-1)=8x-4 | | 5x2-192=0 |