If it's not what You are looking for type in the equation solver your own equation and let us solve it.
8n^2+41n+5=0
a = 8; b = 41; c = +5;
Δ = b2-4ac
Δ = 412-4·8·5
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-39}{2*8}=\frac{-80}{16} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+39}{2*8}=\frac{-2}{16} =-1/8 $
| 2(3b-11)=8b-11 | | 20x+60=20 | | 33.3(2x-4)+5=-66.6(x+1) | | 12x+3=4x+7 | | 9.1-4w=9 | | -4x-4+5x=-6+90 | | 24-p=13 | | 7x+3(x-2)=4(x+1)-3 | | 7+3x-x2=0 | | (5/p+3)+(2/p+3)=0 | | 27+12t-t^2=0 | | (6^(3m)*10^m)/(30^m)=2^(3m)3^(2m) | | X-1÷x-2+x-3÷x-4=10÷3 | | 32=8j | | 9x+2(2x-15)=-6 | | t(2t-4)=t^2+3t+2 | | 14x-6+8x-8=68 | | 3r^2+14r-49=0 | | (x+3)^2=21 | | t(2t-4)=t°+3t+2 | | (x-3)^2=21 | | 15x+35=5 | | (x-3)^2=3 | | 5/p=3 | | b+23/10= 3 | | (x+3)^2=3 | | 3(2y-1)=6y-4 | | 4y-5(4y-8)-12=26 | | 70/100=x/3 | | x^2-1/x+5=0 | | 5x+1=9x+3 | | 4-(x^2-2x+5)=-4 |