If it's not what You are looking for type in the equation solver your own equation and let us solve it.
x^2+40x-48=0
a = 1; b = 40; c = -48;
Δ = b2-4ac
Δ = 402-4·1·(-48)
Δ = 1792
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{7}}{2*1}=\frac{-40-16\sqrt{7}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{7}}{2*1}=\frac{-40+16\sqrt{7}}{2} $
| 2x+1+×=6-(×-1) | | 4x-7+6x-15-13x=8 | | 8n=4n+20 | | √(y+7)+5=y | | 0.5x+45=3x | | 2x2-8×=64 | | F(x)=1(x+3)^2+1 | | -2(4x-5=10 | | 6x-78=267.8 | | 72000=3000(1+.05)*x | | (x-20)/x=6.67 | | -4y+30=80 | | 4.5+10m=7.63 | | b(9-2b)=26 | | 17x+6=18x | | 1-0.75(2+v)=-5 | | (4+r)/(2+3r)=0 | | X=(4+x)/(2+3x) | | 2x-1(/3)=5 | | 16=r/13 | | 16-x=-11+3x | | X=(4+r)/2+3r | | |-12|=-9+x | | P+1/p-2=p+1/p-3 | | x+-12+3x=-2(x+18) | | 8.5=n+3.7 | | m/10=-7 | | -2(2+6x)=7(-1+6x)+3 | | 5x-3(x-3)=-9+5x+6 | | 2.3+10m=7.54 | | -3=(w+2)+4w | | -6(1-2x)-5(x+6)=-8 |