If it's not what You are looking for type in the equation solver your own equation and let us solve it.
16x^2-41x-25=0
a = 16; b = -41; c = -25;
Δ = b2-4ac
Δ = -412-4·16·(-25)
Δ = 3281
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-\sqrt{3281}}{2*16}=\frac{41-\sqrt{3281}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{3281}}{2*16}=\frac{41+\sqrt{3281}}{32} $
| 2(s-4)=10 | | 3.4x+7.1=2.5x+8 | | 50+30x=320 | | 2r+4r=10 | | 10x+12=2x | | 3q-5=10-2q | | 9x–10=100 | | 3x/4-3=15 | | 14x+28=2x | | x^2+20x+250=0 | | (2p)/3-(p+1)/6=2 | | 2p/3-(p+1)/6=2 | | 3x/4-3=16 | | 5x9=10x | | 3x-4=0.5(x+12) | | 5x–8=47 | | 5(7-2h)=1/2h-3 | | 3p+4=10p | | 3x-4=1/2(x+12) | | 20=x/7+2 | | 3t+4=t+20 | | m÷5-2=6 | | x=1/2(12+(3x-4)) | | x+1/2x+3=3/8 | | 8^(x+1)=16 | | 22=5x2 | | 12χ²+x-6=0 | | 12x²+x=6 | | 2x+3=-0,5x+1 | | 36-y=4y | | 15w+4=24-7w | | (x+6)/10=7/5+(x-1)/7 |