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(2x+3)(x+10)=230
We move all terms to the left:
(2x+3)(x+10)-(230)=0
We multiply parentheses ..
(+2x^2+20x+3x+30)-230=0
We get rid of parentheses
2x^2+20x+3x+30-230=0
We add all the numbers together, and all the variables
2x^2+23x-200=0
a = 2; b = 23; c = -200;
Δ = b2-4ac
Δ = 232-4·2·(-200)
Δ = 2129
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{-(23)-\sqrt{2129}}{2*2}=\frac{-23-\sqrt{2129}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{2129}}{2*2}=\frac{-23+\sqrt{2129}}{4} $
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