If it's not what You are looking for type in the equation solver your own equation and let us solve it.
y^2-15=8y
We move all terms to the left:
y^2-15-(8y)=0
a = 1; b = -8; c = -15;
Δ = b2-4ac
Δ = -82-4·1·(-15)
Δ = 124
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{31}}{2*1}=\frac{8-2\sqrt{31}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{31}}{2*1}=\frac{8+2\sqrt{31}}{2} $
| x+(x*0,1)=247,5 | | 2x-45=x+45 | | 7k-4k÷k=-(-2k) | | 7k-4k=÷k-(-2k) | | (5/10)*(x+40)=30 | | 12x-3=x+1 | | 11b=4b+98 | | (3-i)(4+7i)+3i(1-i)-1=0 | | 4m+3/3=5 | | 4(2w-1)=7w+1 | | (5+3i)(3+5i)=0 | | 8*9^(x-3)+4^(x-3)=3^(2x-4) | | (4+i)+(2-i)-(1-i)=0 | | H(x)=5x2-30x+30 | | (-1+2i)-(4-3i)=0 | | (2+5i)+(4+3i)=0 | | 42-x=-42 | | Q=60-0,3p | | 4x= 14x | | Q=30+0,2p | | X/1+3/x=13/2 | | (x-7)/(2)=(-3)/(x-2) | | 4(3x-2)=2-5 | | 8+4i/5=10 | | 90x=10 | | 20=3+13b/7 | | 4x=+7 | | 5x+3=8x4 | | (3x+4)=(4x+3) | | 5X-2=|13x-26| | | 0=-16t^2+29t-11 | | 0=-16t^2+24t-11 |