If it's not what You are looking for type in the equation solver your own equation and let us solve it.
3k^2-17=0
a = 3; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·3·(-17)
Δ = 204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*3}=\frac{0-2\sqrt{51}}{6} =-\frac{2\sqrt{51}}{6} =-\frac{\sqrt{51}}{3} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*3}=\frac{0+2\sqrt{51}}{6} =\frac{2\sqrt{51}}{6} =\frac{\sqrt{51}}{3} $
| 5x+8=3(x-4)+6 | | 6m•4=24 | | 6-8(2x-4)=-58 | | 5(2x-1)=-x+28 | | 9x.3+.4=3.5 | | 3p–4=8. | | P=120-3q | | 4(x-3)=5x-9 | | 4x-23=4(5x+5)+5 | | 5.1a=12 | | 6x+22=4x+26 | | (X+2)2+(2x-1)2=5x(x+1) | | 4b-23=22 | | 4(47-x)+2x=124 | | 5x/11=7 | | 3(4x+3)2=48 | | 20-5x=6-2x | | –2t−4=2−2t | | 1x²+20x-11=-1 | | 6z+3-4z=9) | | 40-x=x+14 | | 3(4x+3)^2=48 | | x²+20x-11=-1 | | 50+4x=90-4x | | 16.x²+20x-11=-1 | | -6=0.75(r�12) | | (14t-2)-(6t+15)=-9 | | 4b-23=-22 | | 78+b=56 | | 3n+46=10 | | 12/x=36/24 | | X^2+10x+16=0. |