If it's not what You are looking for type in the equation solver your own equation and let us solve it.
x^2+24x-64=0
a = 1; b = 24; c = -64;
Δ = b2-4ac
Δ = 242-4·1·(-64)
Δ = 832
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{13}}{2*1}=\frac{-24-8\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{13}}{2*1}=\frac{-24+8\sqrt{13}}{2} $
| x2=0 | | 4=-2x+25 | | 5(3x-5)=7 | | x^2+10x-169=0 | | 3x-10º=4x+10º | | 400=112×14×r | | (4w+3)=(20-6w) | | 25=2x+13 | | 5(x-2=-30 | | 2x^2-3x-20=x^2+34 | | 0.06x+x=150 | | -5-10x=20 | | 7(x+1=2x+57 | | x^2+2x=40-x | | 4x-1=17-5x | | 5*(x+1)-1=3*(3x-4) | | 6(x+2)=8x-6 | | 3*(x+2)=10+2*(5-2x) | | 12d+15=11 | | 2k*2k=k^2 | | X^3+2x^2-0.5x+5=0 | | 4*(2x-1)-(3x+1)=2*(x-4) | | 4x-7=x+18-x | | 4*(2x-1)-(3x+1)=2*8(x-4) | | £20=a-£225 | | 2x+5*(x+1)=3*(x+3) | | x/12-3=7 | | 5(3x+20-4x=87 | | 3x2+12x−135=0 | | 5(3x+2)-3x=87 | | 8x-14=14x | | 3*(4x-1)=4*(2x+3) |