If it's not what You are looking for type in the equation solver your own equation and let us solve it.
y^2=12y=36
We move all terms to the left:
y^2-(12y)=0
a = 1; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·1·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*1}=\frac{0}{2} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*1}=\frac{24}{2} =12 $
| x/2-15=25 | | 4x-28=3x+1 | | 3x+4x+4=-19+2 | | 1=1/3z | | 1/3x+1=5-x | | 47-2x=5-2x | | 110=1/2h(2+9) | | 8y=5y-24 | | 2x/7=x-1/3 | | 2x+4=5x-8= | | 4y=12y-48 | | (x+2)(x-5)=(x+1)(x+1)-10 | | -2k-15=6 | | 6x-22=2x-6 | | 6u+6=4 | | 8/(3x)+3/(4x)=0 | | 16(3x-9)/6=8 | | 3(2x−1)+7=6x−3−2(2−4) | | 4x+4=x16 | | (4x+6)=(12x-18) | | -3x-8=x+6 | | 3(2x-3)=2x-3 | | 5−14x=15−14x | | 8/3x+3/4x=0 | | 3x+2x+6x=33 | | 31(g−3)g−3g=3 | | |2t|=6 | | 3(9+4x−5)=3(9+4x−5) | | 3x+2x+6x=11 | | 3x+5=-8x+10 | | -49+3h=8 | | 129=3(1-6r) |