If it's not what You are looking for type in the equation solver your own equation and let us solve it.
2x^2-3x-39=0
a = 2; b = -3; c = -39;
Δ = b2-4ac
Δ = -32-4·2·(-39)
Δ = 321
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{321}}{2*2}=\frac{3-\sqrt{321}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{321}}{2*2}=\frac{3+\sqrt{321}}{4} $
| 5+2x+3x=180 | | (6x-1)^2-81=0 | | 2x-7x=58-4⋅(3-2) | | 7^2x-1=8^3x+2 | | x+3=-x+3 | | (x+10)(x+14)=(x+14)(×-20) | | X+13=3x-7 | | 3/8n=9/10 | | 2n^2=27-15n | | 9z=88+z | | 5(x^2=2x)=73 | | 2(x+4)=3x+22 | | 5x+44=-17x | | 5-3(x+1)=2(1-x) | | 1/b=0,5 | | 20x+0.12=80 | | 3(x^2+1)-x=2x+9 | | 5^-2x=1/125 | | (y+2)(2√y)=0 | | 5z+2=3z | | x*5+7=32 | | 50-5x(10)=0 | | 50-5x(50-40)=0 | | 50=5x(10) | | -3(m+5)+(4m+2)=8 | | 4-2(x+7=3(x+5) | | 4(2+3y)=56 | | 8m=6m-14 | | 6x+8=4x+46 | | 5-2(3x+4)=18 | | -8a-7+5=-25 | | -6h-5-4h=20 |