If it's not what You are looking for type in the equation solver your own equation and let us solve it.
9x^2+49x=0
a = 9; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·9·0
Δ = 2401
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*9}=\frac{-98}{18} =-5+4/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*9}=\frac{0}{18} =0 $
| Y=-0.06x^2+5 | | 2x²-8=106 | | 0.2x+0.4(4)=0.25(x+4) | | 452.16=4(3.14)r2 | | 0.06y+200+.03y=350 | | 8x=2x+11x | | 3n/5=26 | | 0=2x(x4-24x2-25) | | 3x+7=-34 | | (4x-8)(x-7)=0 | | 0.75x+15=-6 | | u7/2=42 | | 5x–17=-x+7,x= | | x-20-5x=7(3+x)-3x | | (3x-9)^2=2x(2x+29) | | 8x+11=2x-32 | | 2x-2=12-5× | | 28=m20 | | 8w+8=4 | | 8^(6x-10)=5^(4x-10) | | (20x+7)+(5x+3)+(5x+50)=180 | | x-15-3x-10=25 | | 3x+11=2(1+6x) | | 14x-7+x–2+x–43=-16 | | H+h-h+h-H+H-H+h= | | (-16+6b)/2=b | | 70+60(x-1)=490 | | -4+7|-9x+8|=-39 | | 8x+1=−718x+1=−71 | | 3x+21=4x+17 | | 15–2x=4×3 | | 4.9t^2-8.5=0 |