If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5u^2-8u-16=0
a = 5; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·5·(-16)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{6}}{2*5}=\frac{8-8\sqrt{6}}{10} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{6}}{2*5}=\frac{8+8\sqrt{6}}{10} $
| 58=8(4n-6)-7(4-5n) | | 3.2(2x-6)=11.2 | | 4x=2=4x-5 | | 9+3=12+v-8÷2 | | 544+4n=12n | | v/6-v/4=7/6 | | 9+3=12+(v-8)÷2 | | u/4=u/3-2 | | u/3+2/15=u/5 | | -1/2v+6=7 | | x(x+10)=10(-10-x) | | 6=-6y+4(y+3) | | x^2=-(6x+7) | | 3v+8(v-4)=1 | | 14/9=(2x-8)/(x^2-2x-24) | | 2^x+2^(x+2)=40 | | 33=x^2-2x | | z+13/9=40/9 | | 36/s=4;9,10,11 | | Z^2=-(6z+7) | | 4(5x−20)=−20 | | 9h-4=0 | | 5/3+y=2/3 | | 2y=-7(3y+5)+12 | | n-4=2+8n-6n | | 2/5+x=7/5 | | -3+x/2=2 | | 90+11x–20+12x–51=180 | | -22=-8x+8-7x | | 90+x-15+x-15=180 | | 90+x-7+x-7=180 | | j=81 |