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(2x+5)(x-10)=148
We move all terms to the left:
(2x+5)(x-10)-(148)=0
We multiply parentheses ..
(+2x^2-20x+5x-50)-148=0
We get rid of parentheses
2x^2-20x+5x-50-148=0
We add all the numbers together, and all the variables
2x^2-15x-198=0
a = 2; b = -15; c = -198;
Δ = b2-4ac
Δ = -152-4·2·(-198)
Δ = 1809
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1809}=\sqrt{9*201}=\sqrt{9}*\sqrt{201}=3\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{201}}{2*2}=\frac{15-3\sqrt{201}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{201}}{2*2}=\frac{15+3\sqrt{201}}{4} $
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