If it's not what You are looking for type in the equation solver your own equation and let us solve it.
25x^2=5x
We move all terms to the left:
25x^2-(5x)=0
a = 25; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·25·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*25}=\frac{0}{50} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*25}=\frac{10}{50} =1/5 $
| (8t+30)=(-2t-16)=-22 | | 2X-7+6x=-95 | | −2(x+3)=−2(x+1)−4 | | x=−2(x+3)=−2(x+1)−4 | | 17(-3x)=12 | | 5x(2/3+1/2)=10 | | (8t-30)+(-2t-16)=-22 | | -2(2-5)-10=-3(3s+6)-3 | | 5(4r+7)=8r-37 | | 4a+5a^2+2a^2+a^2=0 | | x+2/x=3 | | 41=12d−7 | | 6x+6/8=20 | | (8t+30)=(-2t-16)=22 | | ((x-1)/3)+((x+5)/4))=1/2 | | x(2x+2)=4 | | 7z+5/8=8z+3/8 | | 11+7x=8x+7 | | (8t+30)=(-2t-16)=20 | | 6x+2/3=20 | | x-12=9;3 | | 17x(-3)=12 | | 3/5x+7=12 | | 1-0x=13 | | (8x+33)/2-3x=-(2x-8) | | -7(3+8x)=39+4x | | 2x^2+1=2x+14 | | 2/3d+3/5=2d | | −41a−4=47a−3 | | 3(8x+7)=5(2x+7) | | 7(2n+5)=-13+2n | | .5(2-4x)=2x=13 |