`2x^2 +3x= 20`

solve by completing the square. use the discriminant to find the number of unique real solutions. solve with the quadratic formula.

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`2x^2+3x = 20`

`2x^2+3x-20 = 0`

Discriminant`(Delta)` `= 3^2-4*2*(-20) = 169>0`

Since `Delta > 0` we have two real soluions for the equation.

`x^2+(3x)/2-10 = 0`

`x^2+(3x)/2+9/16-9/16-10 = 0`

`(x+3/4)^2-169/16 = 0`

`(x+3/4)^2 = 169/16`

`(x+3/4) = +-13/4`

`x = +-13/4-3/4`

`x = 5/2` or x = -4

*So the answers are x = 5/2 and x = -4*

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