If it's not what You are looking for type in the equation solver your own equation and let us solve it.
10s^2-41s+4=0
a = 10; b = -41; c = +4;
Δ = b2-4ac
Δ = -412-4·10·4
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-39}{2*10}=\frac{2}{20} =1/10 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+39}{2*10}=\frac{80}{20} =4 $
| 3t^2-23t+14=0 | | 4x+25=6x-23 | | f^2-19f+18=0 | | 9x+15=10x+6 | | 6x+4+7x+1=0 | | 4^(-x+21)=32x | | 10n^2-65n-75=0 | | 4^(-x+21)=326x | | 6j^2-57j-30=0 | | 23w^2-47w=0 | | 48q+9q^2=-64 | | 23y=-5y^2 | | 2z^2+76z=0 | | 4k^2+88k=0 | | 49w^2+10w=0 | | 10s^2+3s-4=0 | | k^2-38k=0 | | k^2-40k=0 | | y=150.000(1.08) | | 3g^2-14g+11=0 | | 3p^2-8p-16=0 | | z^2-42z=0 | | 2(x-5)=6-3x | | 14s^2+25s=0 | | 2^2x-40(2^x)+256=0 | | 10j^2+33j=0 | | 2+2*5+7=x | | 6a-10=84= | | z^2+122z+121=0 | | h^2-144=0 | | 6a-10=84=5 | | 2-2x5=-7 |