Question

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

Answer 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 .