If it's not what You are looking for type in the equation solver your own equation and let us solve it.
y^2-31y+228=0
a = 1; b = -31; c = +228;
Δ = b2-4ac
Δ = -312-4·1·228
Δ = 49
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{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-7}{2*1}=\frac{24}{2} =12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+7}{2*1}=\frac{38}{2} =19 $
| x^2-26x+153=0 | | 5y=2y+22+4 | | 2x+5-13x-5=2 | | 3x+15-4x+17=X+2 | | x*x*x+12x*x+36x+64=0 | | 3a+7+2(3+a)=5a= | | 3a+7+2(3+a)=5a | | 13.33=1/24+1/x | | 16-4*y=0 | | 5k+5=10 | | 400=25v* | | 48=3.14×r | | -4(t+5)=-40 | | 2j=89 | | -6r+7=55 | | u+15/5=2 | | x/4+12=3 | | 16-4x/20=12 | | 18r+8=71 | | w-4/4=-5 | | 3=4a | | 0=(23-5x)/2 | | 3a-16=10 | | ⅕(2+y)=½(y-1) | | 4x−1=x | | 12-9y+5y=8 | | 8/6-y=2 | | p/8=7)p/8=7)p/8=7 | | t-4÷t+3=2÷3 | | 5(z+3)+2(z-4)=3(z-3) | | X²-7x+35=0 | | x•7=-42 |