If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5t^2+t-41=0
a = 5; b = 1; c = -41;
Δ = b2-4ac
Δ = 12-4·5·(-41)
Δ = 821
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{821}}{2*5}=\frac{-1-\sqrt{821}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{821}}{2*5}=\frac{-1+\sqrt{821}}{10} $
| 3x=9x*2−10 | | 3-6(z-2)^3= | | 5(x-4)=3(x/6) | | x^2+16-46=0 | | 2(x+1)-5/3=16 | | 2(x+1)-5/3=16x | | 6(u-1)=-54 | | (5x-6)+(6x)+(8x-8)+(7x)+(4x+14)=600 | | -7=5-2u | | u/9-7=1 | | (2h-5/3)-(h+8/5)=0 | | |2x-3|=0.04 | | 7x+5=4x-14 | | -3c(c-4)-2c-8=9(c+2)+1 | | 5p-2-6p=9 | | v/3-12=19 | | 18(x-3)=54 | | 16y2+16y+4=(y)2 | | 2(2q+1.5=18-q | | 31q+18=328 | | 5^(3x+4)=40 | | 5^3x+4=40 | | -12x3+-36x=0 | | 10y=10,000,000 | | 2-6k=20 | | -17m+2=70 | | -5=43-8y | | |2x-9|=-13 | | x+6x=700 | | 5x-9/2x=1 | | t^+19t+90=0 | | 5=-3y+29 |