If it's not what You are looking for type in the equation solver your own equation and let us solve it.
x^2+46x-360=0
a = 1; b = 46; c = -360;
Δ = b2-4ac
Δ = 462-4·1·(-360)
Δ = 3556
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3556}=\sqrt{4*889}=\sqrt{4}*\sqrt{889}=2\sqrt{889}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-2\sqrt{889}}{2*1}=\frac{-46-2\sqrt{889}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+2\sqrt{889}}{2*1}=\frac{-46+2\sqrt{889}}{2} $
| 4-3x-7x=-8 | | 5^3x+2=25^x-8 | | 3(4+5y)+15y=-1 | | 3x-15=27+5 | | X^3-8x^2+4x-60=0 | | x(x^2-3x+3)=0.05 | | 2{-4x-13}=37+13x | | −10s=5 | | 2(4z-4)=32 | | 7x-35-4x=51+27 | | 4r^2+12+10=0 | | 15x-8+13x=x | | –3+4x–(1+3x)=1–2x+4 | | 9(x+5)=15 | | −20s=15 | | (x+2)(x+2)=(x-5)(x-5)+7 | | (x+2)2=(x-5)2+7 | | H=-16t^2+55+4 | | 4t=39−20t | | 6y+6=2y+9 | | 7+m/8=-22 | | 15x+(-8)+13x=x | | 188+90+2y=180 | | 0.01^(x)=100 | | -2x-1/4=6+3x | | (4x)+1=x+(-4) | | 4x*20=168 | | 6x*20=168 | | a/4=(10/2) | | 4x+20=168 | | 12n-24=-14-28 | | (2n+4)+6=-9+4(2n+1) |