How do you solve ... 0=-5t^2+20t+1

There were 2 solutions that i came up with. Here is the first one. I rearranged the eqaution by subtracting what is to the right of thr equal sign. Multiply the coefficeint of the first term by the constant 5 x (-1) = - 5. Then you would find 2 factors of - 5 whose sum equals the coefficient of the middle term which is 20. - 5+1 = - 4 and - 1+5=4. the eqaution then comes to 5t^2-20t-1=0. you would solve 5t^2-20t-1=0. you would divide both sides of the equal sign by 5. t^-4t-1/5=0. then t^2-4t=1/5. add 4 to both sides of the equation. so we get 21/5 + 4 + 4t - t^2=21/5. it then comes out to be t^2-4t+t=t-2^2. according to the law of transitivity it is t-2^2=21/5.

t = (20+√420) / 10=2+1/5√ 105 = 4.049.