If it's not what You are looking for type in the equation solver your own equation and let us solve it.
2x^2-40x+13=0
a = 2; b = -40; c = +13;
Δ = b2-4ac
Δ = -402-4·2·13
Δ = 1496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1496}=\sqrt{4*374}=\sqrt{4}*\sqrt{374}=2\sqrt{374}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{374}}{2*2}=\frac{40-2\sqrt{374}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{374}}{2*2}=\frac{40+2\sqrt{374}}{4} $
| 1/2n+10=1/4n+54 | | -2(-w+2w-3)=30 | | d2=289 | | (x+4)/6+(x-5)/12=14 | | 17=13-6×a | | 16x-10+9=99 | | 3/7=t/4 | | 37=t4 | | s2=196 | | 20x+1,800=20x | | (x+4)/(6)+(x-5)/(12)=14 | | -8+g=-11 | | 7y+7=6-Y | | (3x+9)+(3x+8)+(3x)=(3x)+(3x)+(4x+2) | | -9(t-2)=2(t+9) | | 15=-6+3x | | q=0.3 | | 4^2-4x=17 | | 6(8x+5)=-3(2-4x) | | 5x/9=-5 | | 4-t=3(+-1)-5 | | 2=x-7/0.10 | | 2*4-1=3*4w+2 | | 16=3a+4=-3 | | 9+4z=45 | | 4x+15+2x+3=180 | | 6xx5+5=53 | | Q(x)=-5x+6 | | 16=3a+4=1 | | y-3=0.5 | | 9x/6=-1 | | X^2+3=(x+4)(x-5) |