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(x+3)(2x-5)-1404=0
We multiply parentheses ..
(+2x^2-5x+6x-15)-1404=0
We get rid of parentheses
2x^2-5x+6x-15-1404=0
We add all the numbers together, and all the variables
2x^2+x-1419=0
a = 2; b = 1; c = -1419;
Δ = b2-4ac
Δ = 12-4·2·(-1419)
Δ = 11353
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{11353}}{2*2}=\frac{-1-\sqrt{11353}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{11353}}{2*2}=\frac{-1+\sqrt{11353}}{4} $
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