If it's not what You are looking for type in the equation solver your own equation and let us solve it.
b^2-2b-51=0
a = 1; b = -2; c = -51;
Δ = b2-4ac
Δ = -22-4·1·(-51)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{13}}{2*1}=\frac{2-4\sqrt{13}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{13}}{2*1}=\frac{2+4\sqrt{13}}{2} $
| 5x-7+4x+18=360 | | -3(-2)=4y+3 | | -9x-15+x-2=47 | | -3x-23=111 | | x-16+70=180 | | -3x=4(-3/4)+3 | | 2x+1564x+192=x | | x÷4-8=2 | | −9x-15+x−2=47 | | v^2-6v-55=0 | | 4x+16+100=180 | | 45=3u-12 | | 2(5x+2)-4x=-2 | | 10x-(5+4x)=14x-5 | | 3x+9-x=6+2x+3 | | -2z+10+2z=16z+7 | | 8y+2.5=105 | | -2z+10+7z=162+7 | | 5(v+3)=9v+27 | | 5x-8x-3x=10 | | 2(5x-3)+2x=4x-22 | | 6-4x+6x=30 | | (y-2)^2=2 | | 4=(-3r-5) | | -5+7x=-2x-5 | | 7x+12+3x+5x+2=180 | | -29=5/9(f—32) | | -14(z-5)=-+ | | 5+7r+3r=5r-4 | | 1-2p+5=5p+6 | | s*6+23=155 | | 3(x-3)-17=4 |