How do you solve #T^2+4T-45=0#?

I have spotted that #5xx9=45# and that #9-5=4# so it looks as though we have found the factors.

So #(T+?)(T+?)-> (T+9)(T-5)#

As the constant is negative that means one of the {9,5} has to be positive and the other negative so that their product is negative.

I have made the 9 positive as it is the larger of the two numbers and we need their difference to be positive.

Lets test it by multiplying out the guessed at bracket structure.

#color(blue)((T+9)(T-5))color(brown)( -> T^2-5T+9T-45)color(green)( = T^2+4T-45)#
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Now we link this back to the question:

#T^2+4T-45=0#

so

#(T+9)(T-5)=0#

This can only be true if T=-9 or T=+5

#T^2# + 4T - 45 = 0

(T + 9)(T - 5) = 0

T = -9 et 5

So, this equation's solutions are -9 and 5.

To solve quadratic equations, you always have to put all the numbers and variables to one side of the equation, so that the other side equals 0.

Here are a few exercises for your practice:

1. Solve each equation by simplifying and then factoring:

#2x^2# + 21x + 34 = 0

#4x^2# + 6 = 14x

#(3x + 4) / 11# = #(5x - 6) / (2x)#

Hopefully you understand now!